Chapter 7 Integrals (Concepts)

Introduction to Integrals

In the study of calculus, we have primarily focused on differentiation, which deals with finding the rate of change of a function. Now, we embark on the study of integration, which is essentially the reverse or inverse process of differentiation. Just as subtraction is the inverse of addition, or division is the inverse of multiplication, integration is the inverse operation to differentiation. If we are given a function, differentiation tells us how fast it is changing. Conversely, if we know the rate of change of a function (i.e., its derivative), integration allows us to find the original function. This inverse process is called anti-differentiation.

Besides being the inverse of differentiation, integration has a profound connection to the concept of finding the area under a curve. Calculating areas of irregular shapes using geometric formulas can be impossible. Integration provides a powerful method to determine the area of a region bounded by a curve, the x-axis (or y-axis), and vertical (or horizontal) lines. This connection is formalised through the definition of the definite integral.

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Antiderivative (or Primitive)

The process of finding a function whose derivative is a given function is called anti-differentiation. The function we find is called an antiderivative.

Definition: A function $F(x)$ is called an antiderivative or primitive of a function $f(x)$ on an interval $I$ if the derivative of $F(x)$ with respect to $x$ is equal to $f(x)$ for all values of $x$ in the interval $I$.

Let's consider an example. We know that the derivative of $x^2$ is $2x$. So, $F(x) = x^2$ is an antiderivative of $f(x) = 2x$. What about the derivative of $x^2 + 5$? It is $\frac{d}{dx}(x^2 + 5) = 2x + 0 = 2x$. So, $x^2 + 5$ is also an antiderivative of $2x$. What about the derivative of $x^2 - 100$? It is $\frac{d}{dx}(x^2 - 100) = 2x - 0 = 2x$. So, $x^2 - 100$ is also an antiderivative of $2x$.

In general, if $F(x)$ is an antiderivative of $f(x)$, then for any arbitrary constant $C$, the function $G(x) = F(x) + C$ is also an antiderivative of $f(x)$, because:

Thus, if a function $f(x)$ has one antiderivative $F(x)$, it actually has an entire family of antiderivatives, given by $F(x) + C$, where $C$ can be any real number. This constant $C$ is called the constant of integration.

This collection of all antiderivatives of $f(x)$ is called the indefinite integral of $f(x)$.

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Two Types of Integrals

Calculus deals with two main types of integrals, each serving a different purpose:

1. Indefinite Integral: The indefinite integral of a function $f(x)$, denoted by $\int f(x) dx$, represents the family of all antiderivatives of $f(x)$. If $F(x)$ is one antiderivative of $f(x)$, then the indefinite integral is written as $\int f(x) dx = F(x) + C$, where $C$ is the arbitrary constant of integration. The term 'indefinite' refers to the fact that the result is not a single function but a collection of functions differing by a constant.

2. Definite Integral: The definite integral of a function $f(x)$ over a specific interval from $a$ to $b$, denoted by $\int_{a}^{b} f(x) dx$, represents a specific numerical value. Geometrically, if $f(x) \geq 0$ on $[a, b]$, this value corresponds to the area of the region bounded by the curve $y=f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$. If $f(x)$ takes negative values, the definite integral represents the signed area (areas below the x-axis are considered negative). The numbers $a$ and $b$ are called the lower and upper limits of integration, respectively.

The profound relationship between the indefinite integral (antidifferentiation) and the definite integral (area under the curve) is established by the Fundamental Theorem of Integral Calculus, which we will study later. This theorem is a cornerstone of calculus, linking the two seemingly different concepts.

Indefinite Integrals

As discussed in the introduction, the process of finding antiderivatives is called integration. When we find all possible antiderivatives of a function, we are finding its indefinite integral. The indefinite integral of a function $f(x)$ is the entire collection or family of all functions $F(x)$ whose derivative is $f(x)$.

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Notation of Indefinite Integral

The indefinite integral of a function $f(x)$ with respect to the variable $x$ is denoted by the symbol $\int f(x) dx$.

If $F(x)$ is any one antiderivative of $f(x)$ (meaning $F'(x) = f(x)$), then the indefinite integral of $f(x)$ is given by:

Here's a breakdown of the notation:

• $\int$: This elongated 'S' symbol is the integral sign, introduced by Leibniz. It represents the operation of integration.
• $f(x)$: This is the function being integrated. It is called the integrand.
• $dx$: This indicates that the integration is being performed with respect to the variable $x$. This is crucial, especially when dealing with functions of multiple variables or when performing integration by substitution.
• $F(x)$: This is any one function whose derivative is $f(x)$, i.e., an antiderivative of $f(x)$.
• $C$: This is the constant of integration. It represents the entire family of antiderivatives. Since the derivative of any constant is zero, adding an arbitrary constant $C$ to an antiderivative $F(x)$ results in another antiderivative $F(x) + C$. The value of $C$ is not determined in an indefinite integral, hence the term "indefinite".

The process of evaluating or finding an indefinite integral is also referred to as integration.

Indefinite integration is the inverse operation to differentiation. This relationship can be expressed symbolically:

1. If we differentiate an indefinite integral, we get back the original function:

For example, $\frac{d}{dx}(\int 2x dx) = \frac{d}{dx}(x^2 + C) = 2x + 0 = 2x$, which is the original function.

2. If we integrate the derivative of a function, we get the original function back, but with an arbitrary constant of integration:

For example, $\int \frac{d}{dx}(x^3) dx = \int 3x^2 dx$. We know that the antiderivative of $3x^2$ is $x^3$, so $\int 3x^2 dx = x^3 + C$. This gives us the original function $x^3$ plus the constant of integration.

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Geometrical Interpretation of Indefinite Integral

The indefinite integral $\int f(x) dx = F(x) + C$ represents not a single curve, but a family of parallel curves. Each different value of the constant of integration $C$ corresponds to a different curve in this family.

Consider the example $f(x) = 2x$. The indefinite integral is $\int 2x dx = x^2 + C$. The general form of the antiderivative is $y = x^2 + C$. By assigning different real values to $C$, we get different parabolas:

• If $C=0$, we get $y = x^2$.
• If $C=1$, we get $y = x^2 + 1$.
• If $C=-2$, we get $y = x^2 - 2$.
• If $C=5.5$, we get $y = x^2 + 5.5$.
• And so on.

These curves are vertical shifts of the basic curve $y = F(x)$ (in this case, $y=x^2$).

The key property of this family of curves is that at any given value of $x$, the slope of the tangent line to each curve is the same and is equal to $f(x)$. For example, at $x=1$, the slope of the tangent to any curve $y = x^2 + C$ is $\frac{dy}{dx} = 2x = 2(1) = 2$. All curves in the family $y=x^2+C$ have a tangent with a slope of 2 at $x=1$. This is true for any specific $x$ value – the tangents at points with the same x-coordinate on different curves in the family are parallel.

The constant of integration $C$ thus represents the degree of freedom we have when finding an antiderivative; it indicates that there are infinitely many antiderivatives for a given function $f(x)$, all related by a vertical shift. To find a specific antiderivative (and thus determine the value of $C$), we need additional information, usually in the form of an initial condition (a point $(x_0, y_0)$ that the curve passes through), which is common in solving differential equations.

Properties of Indefinite Integrals

The properties of indefinite integrals are directly inherited from the corresponding properties of derivatives. These properties make the process of integration practical, allowing us to integrate sums and differences of functions, and constant multiples of functions, by integrating each term separately. These are often referred to as the linearity properties of integration.

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Linearity Properties

Let $f(x)$ and $g(x)$ be two integrable functions on an interval, and let $k$ be a constant. The following properties hold for indefinite integrals:

1. Integral of Sum or Difference: The indefinite integral of the sum or difference of two functions is equal to the sum or difference of their individual indefinite integrals.

This property means we can integrate term by term when dealing with sums or differences of functions.

2. Integral of a Constant Multiple: The indefinite integral of a constant times a function is equal to the constant times the indefinite integral of the function.

This property allows us to factor out constant coefficients before integrating.

These two properties can be combined into a single general linearity property for a finite number of functions:

where $c_1, c_2, \dots, c_n$ are constants and $f_1, f_2, \dots, f_n$ are integrable functions.

Derivation of Linearity Properties

The proofs of these properties rely directly on the definition of an indefinite integral and the properties of differentiation.

Derivation of Property 1: $\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx$

Let $F(x)$ be an antiderivative of $f(x)$, so $\frac{d}{dx}(F(x)) = f(x)$. Let $G(x)$ be an antiderivative of $g(x)$, so $\frac{d}{dx}(G(x)) = g(x)$.

Consider the derivative of the sum/difference of these antiderivatives, $F(x) \pm G(x)$, using the sum/difference rule of differentiation:

By the definition of an indefinite integral, if the derivative of a function is $f(x) \pm g(x)$, then that function is an antiderivative of $f(x) \pm g(x)$. Thus, $F(x) \pm G(x)$ is an antiderivative of $f(x) \pm g(x)$.

The indefinite integral of $f(x) \pm g(x)$ is the family of all its antiderivatives:

Now consider the right side of the property: $\int f(x) dx \pm \int g(x) dx$.

By definition, $\int f(x) dx = F(x) + C_1$ and $\int g(x) dx = G(x) + C_2$, where $C_1$ and $C_2$ are arbitrary constants.

Since $C_1$ and $C_2$ are arbitrary constants, their sum or difference ($C_1 \pm C_2$) is also an arbitrary constant. Let $C = C_1 \pm C_2$.

Comparing equations (A) and (B), we see that $\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx$.

Derivation of Property 2: $\int k f(x) dx = k \int f(x) dx$

Let $F(x)$ be an antiderivative of $f(x)$, so $\frac{d}{dx}(F(x)) = f(x)$. Let $k$ be a constant.

Consider the derivative of $k F(x)$, using the constant multiple rule of differentiation:

By the definition of an indefinite integral, if the derivative of a function is $k f(x)$, then that function is an antiderivative of $k f(x)$. Thus, $k F(x)$ is an antiderivative of $k f(x)$.

The indefinite integral of $k f(x)$ is the family of all its antiderivatives:

Now consider the right side of the property: $k \int f(x) dx$. By definition, $\int f(x) dx = F(x) + C_1$, where $C_1$ is an arbitrary constant.

If $k \neq 0$, then $k C_1$ is also an arbitrary constant. Let $C = k C_1$.

Comparing equations (C) and (D), we see that $\int k f(x) dx = k \int f(x) dx$. If $k=0$, then $\int 0 \cdot f(x) dx = \int 0 dx = C'$ (an arbitrary constant). And $0 \cdot \int f(x) dx = 0 \cdot (F(x) + C_1) = 0$. For the property to hold in this case, we interpret $0 \int f(x) dx$ as the constant function zero plus an arbitrary constant, i.e., $0+C'$. So, the property holds for $k=0$ as well.

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Integration by Substitution

While we can find the antiderivatives of basic functions using standard formulas, many integrals that appear in practice are more complex and cannot be evaluated directly. One of the most fundamental and widely used techniques for evaluating such integrals is the method of integration by substitution, also known as the change of variable method or u-substitution. This method simplifies the integrand by replacing the independent variable with a new variable, effectively transforming the integral into a form that can be solved using standard integration formulas. The method is essentially the inverse process of the Chain Rule of differentiation.

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The Method of Substitution

The method of substitution is based on the idea of reversing the Chain Rule. Recall the Chain Rule for differentiation: if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. This means that the antiderivative of $f'(g(x)) \cdot g'(x)$ is $f(g(x))$.

The substitution method makes this process more systematic. It involves making a substitution $u = g(x)$. Then, the differential $du$ is related to $dx$ by the derivative of the substitution:

Multiplying by $dx$ (treating $du$ and $dx$ as differentials), we get:

Now consider an integral of the form $\int h(x) dx$. If we can rewrite the integrand $h(x)$ in the form $f(g(x)) g'(x)$, we can make the substitution $u = g(x)$ and $du = g'(x) dx$.

The integral then transforms from an integral with respect to $x$ to an integral with respect to $u$:

The goal is to choose the substitution $u = g(x)$ such that the resulting integral $\int f(u) du$ is simpler and can be evaluated using known integration formulas. Once the integral with respect to $u$ is evaluated, we substitute back $u = g(x)$ to express the result in terms of the original variable $x$.

If $\int f(u) du = F(u) + C$, then

The success of this method depends heavily on choosing the correct substitution. A good substitution $u = g(x)$ is usually a part of the integrand whose derivative $g'(x)$ is also present as a factor in the integrand (possibly multiplied by a constant).

Guidelines for Choosing Substitution

Look for expressions within the integrand that could be a function $g(x)$, such that its derivative $g'(x)$ (or a constant multiple of it) is also present in the integrand. Common patterns include:

1. Inner Function of a Composite Function: If the integrand contains a composite function $f(g(x))$, try substituting $u = g(x)$. You will need $g'(x) dx$ to be present.

• Integral form: $\int f(ax+b) dx$. Substitution: $u = ax+b$. Then $du = a dx$, so $dx = \frac{1}{a} du$. The integral becomes $\int f(u) \frac{1}{a} du = \frac{1}{a} \int f(u) du$.
• Integral form: $\int (g(x))^n g'(x) dx$. Substitution: $u = g(x)$. Then $du = g'(x) dx$. The integral becomes $\int u^n du$.

2. Expression in the Denominator: If the integrand is a fraction where the numerator is the derivative of the denominator (or a constant multiple), try substituting $u$ as the denominator.

• Integral form: $\int \frac{g'(x)}{g(x)} dx$. Substitution: $u = g(x)$. Then $du = g'(x) dx$. The integral becomes $\int \frac{du}{u} = \ln|u| + C = \ln|g(x)| + C$.

3. Expression under a Root or Power: If there is an expression raised to a power or under a square root, try substituting $u$ equal to that expression.

• Integral form: $\int x^n \sqrt{ax^{n+1} + b} dx$. Substitution: $u = ax^{n+1} + b$. Then $du = a(n+1)x^n dx$. You would need $x^n dx$ present.

The key is that after substitution, the integral should contain only the new variable $u$ and $du$. There should be no $x$ terms remaining after the substitution.

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Standard Integrals

Just as there are fundamental formulas for differentiating basic functions, there are also a set of standard formulas for integrating basic functions. These are derived directly from the definition of the indefinite integral (i.e., by reversing standard differentiation formulas) or by using simple substitution. These standard integrals form the building blocks for evaluating more complex integrals using techniques like substitution, integration by parts, partial fractions, etc. A good understanding and memorization of these formulas are essential for success in integral calculus.

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Basic Integration Formulas

Here is a list of fundamental indefinite integral formulas:

1. Power Rule for Integration:

(Derived from $\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) = \frac{1}{n+1} \frac{d}{dx}(x^{n+1}) = \frac{1}{n+1} (n+1)x^{(n+1)-1} = x^n$).

2. Integral of $\frac{1}{x}$:

(Derived from $\frac{d}{dx}(\ln|x|) = \frac{1}{x}$ for $x \neq 0$).

3. Integral of Exponential Function $e^x$:

(Derived from $\frac{d}{dx}(e^x) = e^x$).

4. Integral of General Exponential Function $a^x$:

(Derived from $\frac{d}{dx}\left(\frac{a^x}{\ln a}\right) = \frac{1}{\ln a} \frac{d}{dx}(a^x) = \frac{1}{\ln a} (a^x \ln a) = a^x$).

Trigonometric Integrals

The integrals of basic trigonometric functions are derived from the derivatives of the inverse trigonometric functions.

5. $\int \cos x dx = \sin x + C$ (Since $\frac{d}{dx}(\sin x) = \cos x$)

6. $\int \sin x dx = -\cos x + C$ (Since $\frac{d}{dx}(-\cos x) = -(-\sin x) = \sin x$)

7. $\int \sec^2 x dx = \tan x + C$ (Since $\frac{d}{dx}(\tan x) = \sec^2 x$)

8. $\int \text{cosec}^2 x dx = -\cot x + C$ (Since $\frac{d}{dx}(-\cot x) = -(-\text{cosec}^2 x) = \text{cosec}^2 x$)

9. $\int \sec x \tan x dx = \sec x + C$ (Since $\frac{d}{dx}(\sec x) = \sec x \tan x$)

10. $\int \text{cosec } x \cot x dx = -\text{cosec } x + C$ (Since $\frac{d}{dx}(-\text{cosec } x) = -(-\text{cosec } x \cot x) = \text{cosec } x \cot x$)

The integrals of $\tan x, \cot x, \sec x, \text{cosec } x$ can be found using substitution:

11. $\int \tan x dx = \int \frac{\sin x}{\cos x} dx$. Let $u = \cos x$, then $du = -\sin x dx$, so $\sin x dx = -du$.

Using logarithm properties, $-\ln|\cos x| = \ln(|\cos x|^{-1}) = \ln\left|\frac{1}{\cos x}\right| = \ln|\sec x|$. So, $\int \tan x dx = \ln|\sec x| + C$.

12. $\int \cot x dx = \int \frac{\cos x}{\sin x} dx$. Let $u = \sin x$, then $du = \cos x dx$.

So, $\int \cot x dx = \ln|\sin x| + C$.

13. $\int \sec x dx$: This integral is not immediately obvious. We can use a trick by multiplying the integrand by $\frac{\sec x + \tan x}{\sec x + \tan x}$.

Now, let $u = \sec x + \tan x$. Then $du = (\frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x)) dx = (\sec x \tan x + \sec^2 x) dx$. The integral becomes $\int \frac{du}{u} = \ln|u| + C$.

14. $\int \text{cosec } x dx$: Similar to $\int \sec x dx$, we multiply by $\frac{\text{cosec } x - \cot x}{\text{cosec } x - \cot x}$.

Let $u = \text{cosec } x - \cot x$. Then $du = (\frac{d}{dx}(\text{cosec } x) - \frac{d}{dx}(\cot x)) dx = (-\text{cosec } x \cot x - (-\text{cosec}^2 x)) dx = (\text{cosec}^2 x - \text{cosec } x \cot x) dx$. The integral becomes $\int \frac{du}{u} = \ln|u| + C$.

An alternative form for $\int \text{cosec } x dx$ using $\text{cosec } x = \frac{1}{\sin x} = \frac{1}{2 \sin(x/2)\cos(x/2)} = \frac{\sec^2(x/2)}{2 \tan(x/2)}$ and substituting $u=\tan(x/2)$ leads to $\ln|\tan(x/2)| + C$.

Inverse Trigonometric Integrals

These formulas are the direct inverse of the differentiation formulas for inverse trigonometric functions.

15. $\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C$, for $-1 < x < 1$. (Since $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$)

16. $\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1} \left(\frac{x}{a}\right) + C$, for $-a < x < a, a > 0$. This can be derived from the previous formula by substituting $x=az$, so $dx = a dz$. $\int \frac{a dz}{\sqrt{a^2-(az)^2}} = \int \frac{a dz}{\sqrt{a^2(1-z^2)}} = \int \frac{a dz}{a\sqrt{1-z^2}} = \int \frac{dz}{\sqrt{1-z^2}} = \sin^{-1} z + C$. Substitute back $z=x/a$.

17. $\int \frac{1}{1+x^2} dx = \tan^{-1} x + C$. (Since $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$)

18. $\int \frac{1}{a^2+x^2} dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C$, for $a \neq 0$. This can be derived from the previous formula by substituting $x=az$, so $dx=a dz$. $\int \frac{a dz}{a^2+(az)^2} = \int \frac{a dz}{a^2(1+z^2)} = \frac{1}{a} \int \frac{dz}{1+z^2} = \frac{1}{a} \tan^{-1} z + C$. Substitute back $z=x/a$.

19. $\int \frac{1}{x\sqrt{x^2-1}} dx = \sec^{-1} x + C$, for $|x| > 1$. (Since $\frac{d}{dx}(\sec^{-1} x) = \frac{1}{|x|\sqrt{x^2-1}}$. Note that for $x>1$, $|x|=x$)

20. $\int \frac{1}{x\sqrt{x^2-a^2}} dx = \frac{1}{a} \sec^{-1} \left(\frac{x}{a}\right) + C$, for $|x| > a, a > 0$. This can be derived from the previous formula by substituting $x=az$, so $dx=a dz$. $\int \frac{a dz}{az\sqrt{(az)^2-a^2}} = \int \frac{a dz}{az\sqrt{a^2(z^2-1)}} = \int \frac{a dz}{az \cdot a\sqrt{z^2-1}} = \frac{1}{a} \int \frac{dz}{z\sqrt{z^2-1}} = \frac{1}{a} \sec^{-1} z + C$. Substitute back $z=x/a$. (Here we assume $x>a>0$, so $z>1$).

Important Note: Remember to always add the constant of integration $C$ to the result of an indefinite integral.

Also note that while the derivatives of $\cos^{-1} x$, $\cot^{-1} x$, $\text{cosec}^{-1} x$ are the negatives of the derivatives of $\sin^{-1} x$, $\tan^{-1} x$, $\sec^{-1} x$ respectively, we typically use the $\sin^{-1}$, $\tan^{-1}$, $\sec^{-1}$ forms as the standard results for these positive integrands. For instance, $\int \frac{-1}{\sqrt{1-x^2}} dx = \cos^{-1} x + C$.

Standard Evaluation

Once we have a list of standard integral formulas and understand the linearity properties, we can start evaluating various indefinite integrals. The simplest integrals can be solved by directly applying the standard formulas after possibly rearranging the integrand using algebraic manipulation or trigonometric identities. For slightly more complex integrals, the method of substitution, along with the standard formulas, is used. This section focuses on the direct application of standard formulas and linearity, sometimes combined with straightforward substitutions.

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Applying Standard Formulas and Linearity

Evaluating an indefinite integral $\int f(x) dx$ using standard formulas and linearity involves the following steps:

1. Examine the integrand $f(x)$.

2. If $f(x)$ is a sum or difference of functions, use the linearity property $\int [g(x) \pm h(x)] dx = \int g(x) dx \pm \int h(x) dx$ to break the integral into simpler integrals.

3. If $f(x)$ contains a constant multiple $k$, use the linearity property $\int k g(x) dx = k \int g(x) dx$ to factor out the constant.

4. Rewrite each resulting integral in a form that matches one of the standard integral formulas. This may involve algebraic manipulation (e.g., expanding expressions, rewriting fractions) or using trigonometric identities.

5. Apply the appropriate standard integral formula to each part.

6. Combine the results and add a single arbitrary constant of integration $C$ at the end (since the sum or difference of arbitrary constants is itself an arbitrary constant).

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Integration of Rational Functions of sin x and cos x

A rational function of $\sin x$ and $\cos x$ is a function that can be expressed as a ratio of two polynomials in $\sin x$ and $\cos x$. Examples include $\frac{\sin x + \cos x}{1 + \sin x \cos x}$, $\frac{1}{1 + \sin x}$, $\frac{\cos x}{2 + \sin x}$, etc. Integrals of such functions can often be challenging to evaluate using standard methods or simple substitutions. However, there exists a universal substitution that transforms any such integral into the integral of a rational function of a new variable, which can then be solved using methods like partial fractions.

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The Universal Substitution $t = \tan(x/2)$

The key to integrating any rational function of $\sin x$ and $\cos x$ is the substitution $t = \tan(x/2)$. This substitution is sometimes called the Weierstrass substitution or the tangent half-angle substitution.

Using this substitution, we can express $\sin x$, $\cos x$, and the differential $dx$ entirely in terms of $t$ and $dt$. We use basic trigonometric identities:

We know that $\sin x = 2 \sin(x/2) \cos(x/2)$ and $\cos x = \cos^2(x/2) - \sin^2(x/2)$. We can rewrite these expressions by dividing the numerator and denominator by $\cos^2(x/2)$ (assuming $\cos(x/2) \neq 0$):

Now, substitute $t = \tan(x/2)$ into these expressions:

Next, we need to find the relationship between $dx$ and $dt$. We differentiate the substitution $t = \tan(x/2)$ with respect to $x$:

Using the chain rule ($\frac{d}{dx}(\tan u) = \sec^2 u \frac{du}{dx}$ with $u=x/2$):

Using the identity $\sec^2 \theta = 1 + \tan^2 \theta$, we have $\sec^2(x/2) = 1 + \tan^2(x/2) = 1 + t^2$:

Rearranging to solve for $dx$ in terms of $dt$:

The three formulas derived above:

• $\sin x = \frac{2t}{1+t^2}$
• $\cos x = \frac{1-t^2}{1+t^2}$
• $dx = \frac{2 dt}{1+t^2}$

allow us to convert any rational function of $\sin x$ and $\cos x$ into a rational function of the single variable $t$. The resulting integral in terms of $t$ can then be evaluated using techniques for integrating rational functions, primarily by decomposing the integrand into partial fractions. This substitution is particularly useful when the integrand contains terms like $\frac{1}{a+b \sin x}$ or $\frac{1}{a+b \cos x}$.

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Integration by Partial Fractions

The method of integration by partial fractions is a powerful technique specifically used to integrate rational functions. A rational function is defined as a function that can be written as the ratio of two polynomials, $P(x)$ and $Q(x)$, where $Q(x)$ is not the zero polynomial. That is, $f(x) = \frac{P(x)}{Q(x)}$.

Rational functions are classified based on the degrees of the polynomials in the numerator and denominator:

• A rational function $\frac{P(x)}{Q(x)}$ is called proper if the degree of the numerator polynomial $P(x)$ is strictly less than the degree of the denominator polynomial $Q(x)$.
• A rational function $\frac{P(x)}{Q(x)}$ is called improper if the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$.

If the rational function is improper, the first step before applying partial fraction decomposition is to perform polynomial long division. This process allows us to write the improper rational function as the sum of a polynomial and a proper rational function:

Here, $S(x)$ is the quotient polynomial (which is easy to integrate using the power rule), and $\frac{R(x)}{Q(x)}$ is the remainder term, which is a proper rational function (the degree of $R(x)$ is less than the degree of $Q(x)$). The focus then shifts to integrating the proper rational function $\frac{R(x)}{Q(x)}$.

The core idea of the partial fraction method is to decompose the proper rational function $\frac{R(x)}{Q(x)}$ into a sum of simpler fractions, called partial fractions, whose integrals are known or can be easily evaluated. The form of this decomposition depends on the factorization of the denominator polynomial $Q(x)$. We need to factorize $Q(x)$ into its linear and irreducible quadratic factors over the real numbers.

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Cases for Partial Fraction Decomposition of a Proper Rational Function $\frac{P(x)}{Q(x)}$

The form of the partial fraction decomposition depends on the nature of the factors of the denominator $Q(x)$:

Case 1: Denominator $Q(x)$ contains non-repeated linear factors.

If $Q(x)$ can be factored into distinct linear factors $(x-a_1), (x-a_2), \dots, (x-a_n)$, where $a_i$ are distinct real numbers, then the proper rational function $\frac{P(x)}{Q(x)}$ can be decomposed as:

where $A_1, A_2, \dots, A_n$ are constants that need to be determined. Each term on the right side is easily integrable: $\int \frac{A}{x-a} dx = A \ln|x-a| + C$.

Case 2: Denominator $Q(x)$ contains repeated linear factors.

If $Q(x)$ contains a linear factor $(x-a)$ repeated $k$ times, i.e., $(x-a)^k$ is a factor of $Q(x)$, then corresponding to this factor, the partial fraction decomposition includes the sum of $k$ fractions:

where $B_1, B_2, \dots, B_k$ are constants. If there are other factors in $Q(x)$, their corresponding partial fractions (as per their type) are added to this sum. The integrals of these terms are of the form $\int \frac{B}{(x-a)^m} dx = B \int (x-a)^{-m} dx$, which can be solved using the power rule for $m \neq 1$, and $\int \frac{B}{x-a} dx = B \ln|x-a| + C$ for $m=1$.

Case 3: Denominator $Q(x)$ contains non-repeated irreducible quadratic factors.

An irreducible quadratic factor is a quadratic expression $ax^2+bx+c$ that cannot be factored into linear factors with real coefficients (i.e., its discriminant $b^2-4ac < 0$). If $Q(x)$ contains such a factor $(ax^2+bx+c)$, then corresponding to this factor, the partial fraction decomposition includes the term:

where $A$ and $B$ are constants. The integral of such a term usually involves completing the square in the denominator and results in logarithmic and inverse tangent functions.

Case 4: Denominator $Q(x)$ contains repeated irreducible quadratic factors.

If $Q(x)$ contains an irreducible quadratic factor $(ax^2+bx+c)$ repeated $k$ times, i.e., $(ax^2+bx+c)^k$ is a factor of $Q(x)$, then corresponding to this factor, the partial fraction decomposition includes the sum of $k$ fractions:

where $A_i$ and $B_i$ are constants. Integration of these terms can be more involved, sometimes requiring trigonometric substitution after completing the square.

Determining the Constants:

After setting up the form of the partial fraction decomposition, the next step is to find the values of the unknown constants ($A_i, B_i$, etc.). This is done by equating the original proper rational function $\frac{P(x)}{Q(x)}$ to its partial fraction decomposition, then clearing the denominators by multiplying both sides by $Q(x)$. This results in an identity between the numerator $P(x)$ and a polynomial involving the unknown constants. The constants can be determined by:

• Substituting specific convenient values of $x$ (especially the roots of $Q(x)$ if they are real).
• Equating the coefficients of like powers of $x$ on both sides of the identity, which leads to a system of linear equations that can be solved for the constants.

Once the constants are found, the original integral is transformed into a sum of simpler integrals, each corresponding to a partial fraction, which can be evaluated using standard formulas.

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Integration by Parts

When we encounter an integral of the product of two functions, i.e., $\int f(x) g(x) dx$, we cannot simply integrate each function separately (unlike differentiation of a product). The technique used for integrating a product of two functions is called Integration by Parts. This method is the integral equivalent of the product rule for differentiation.

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Formula for Integration by Parts

The formula for integration by parts is derived from the product rule of differentiation. Let $u$ and $v$ be two differentiable functions of $x$. The product rule states:

Integrate both sides of this equation with respect to $x$:

Using the property that the integral of a derivative is the function itself (plus a constant, which we'll incorporate later), the left side is simply $uv$. Using the linearity property of integrals on the right side:

Now, let's use differential notation. Let $du = \frac{du}{dx} dx$ and $dv = \frac{dv}{dx} dx$. Substituting these into the equation:

Rearranging this equation to isolate the integral $\int u \, dv$, we get the integration by parts formula:

This is the standard formula for integration by parts.

In the context of integrating a product of two functions $f(x)$ and $g(x)$, we choose one function to be $u = f(x)$ and the remaining part, $g(x) dx$, to be $dv$.

So, we set:

• $u = f(x)$ (the function to be differentiated) $\implies du = f'(x) dx$
• $dv = g(x) dx$ (the function to be integrated) $\implies v = \int g(x) dx$

Substituting these into the formula $\int u \, dv = uv - \int v \, du$, we get another way to write the formula:

The goal of integration by parts is to transform the original integral $\int u \, dv$ into a new integral $\int v \, du$ that is easier to evaluate. The success of this method depends critically on how we choose which part of the integrand is $u$ and which part is $dv$.

Choosing $u$ and $dv$ (ILATE Rule)

The choice of $u$ and $dv$ is crucial. Generally, we choose $u$ as the function that becomes simpler when differentiated, and $dv$ as the remaining part (including $dx$) which is easy to integrate. A helpful mnemonic for deciding which function to choose as $u$ is the **ILATE** (or **LIATE**) rule. This rule suggests the order of preference for choosing $u$ based on the type of function:

• I: Inverse Trigonometric functions (e.g., $\sin^{-1} x, \cos^{-1} x, \tan^{-1} x$)
• L: Logarithmic functions (e.g., $\ln x, \log_a x$)
• A: Algebraic functions (e.g., polynomials like $x^n$, $\sqrt{x}$, etc.)
• T: Trigonometric functions (e.g., $\sin x, \cos x, \tan x$)
• E: Exponential functions (e.g., $e^x, a^x$)

The function type that appears earliest in the ILATE list should generally be chosen as $u$. The remaining part of the integrand, including $dx$, is then assigned to $dv$.

For example, in $\int x e^x dx$, we have an Algebraic function ($x$) and an Exponential function ($e^x$). A comes before E in ILATE, so we choose $u = x$ and $dv = e^x dx$. In $\int x \ln x dx$, we have a Logarithmic function ($\ln x$) and an Algebraic function ($x$). L comes before A, so we choose $u = \ln x$ and $dv = x dx$.

This rule is a guideline, not a strict law, but it works well in most common cases encountered at this level.

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Standard Forms

Beyond the very basic power rule, exponential, logarithmic, and simple trigonometric integrals, there are several other standard integral formulas that are crucial for solving a wide variety of problems. These standard forms often involve rational expressions or square roots where the denominator or the term under the square root is a quadratic expression. These formulas are frequently derived using techniques like trigonometric substitution or completing the square, but they are important enough to be memorized and recognised directly.

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Standard Integrals Involving Quadratic Denominators or Square Roots

These formulas cover integrals where the integrand is of the form $\frac{1}{\text{quadratic}}$ or $\frac{1}{\sqrt{\text{quadratic}}}$. The quadratic expression is typically a simple form like $x^2 \pm a^2$ or $a^2 - x^2$. If the quadratic expression is more general, like $Ax^2+Bx+C$, the first step is usually to complete the square to convert it into one of the standard forms.

1. Integral of $\frac{1}{x^2 - a^2}$:

Derivation using Partial Fractions: We decompose the integrand into partial fractions:

Assume the decomposition is $\frac{A}{x-a} + \frac{B}{x+a}$.

Equating numerators: $1 = A(x+a) + B(x-a)$. Set $x=a$: $1 = A(a+a) + B(a-a) = 2aA \implies A = \frac{1}{2a}$. Set $x=-a$: $1 = A(-a+a) + B(-a-a) = -2aB \implies B = -\frac{1}{2a}$. So, $\frac{1}{x^2-a^2} = \frac{1}{2a(x-a)} - \frac{1}{2a(x+a)}$. Now, integrate:

Using the standard integral $\int \frac{1}{u} du = \ln|u| + C$:

Using logarithm properties ($\ln M - \ln N = \ln(M/N)$):

2. Integral of $\frac{1}{a^2 - x^2}$:

This formula can be derived similarly using partial fractions for $\frac{1}{(a-x)(a+x)}$. Alternatively, note that $\frac{1}{a^2-x^2} = - \frac{1}{x^2-a^2}$. $\int \frac{dx}{a^2-x^2} = - \int \frac{dx}{x^2-a^2} = - \frac{1}{2a} \ln\left|\frac{x-a}{x+a}\right| + C = \frac{1}{2a} \ln\left|\left(\frac{x-a}{x+a}\right)^{-1}\right| + C = \frac{1}{2a} \ln\left|\frac{x+a}{x-a}\right| + C$. For $a^2>x^2$, both $a+x$ and $a-x$ have the same sign, so $\left|\frac{x+a}{x-a}\right| = \left|\frac{a+x}{a-x}\right|$.

3. Integral of $\frac{1}{\sqrt{x^2 - a^2}}$:

4. Integral of $\frac{1}{\sqrt{a^2 - x^2}}$:

This was previously listed as a basic integral, derived from the derivative of $\sin^{-1}(x/a)$.

5. Integral of $\frac{1}{\sqrt{x^2 + a^2}}$:

6. Integral of $\frac{1}{x^2 + a^2}$:

This was also previously listed as a basic integral, derived from the derivative of $\tan^{-1}(x/a)$.

Standard Integrals of the Form $\int \sqrt{\text{quadratic}} dx$

These formulas involve integrals of the square root of a quadratic expression.

7. Integral of $\sqrt{x^2 - a^2}$:

8. Integral of $\sqrt{a^2 - x^2}$:

9. Integral of $\sqrt{x^2 + a^2}$:

These three formulas are often derived using integration by parts (taking $u = \sqrt{\text{quadratic}}$ and $dv=dx$) or using appropriate trigonometric substitutions (e.g., $x=a \sin\theta$ for $\sqrt{a^2-x^2}$, $x=a \tan\theta$ for $\sqrt{a^2+x^2}$, and $x=a \sec\theta$ for $\sqrt{x^2-a^2}$).

Using Standard Forms after Completing the Square

Many integrals involve quadratic expressions that are not in the simple forms $x^2 \pm a^2$ or $a^2 \mp x^2$. In such cases, we use the algebraic technique of completing the square to transform the quadratic expression into one of these standard forms, and then apply the appropriate standard integral formula.

For a general quadratic expression of the form $Ax^2+Bx+C$:

Factor out $A$: $A(x^2 + \frac{B}{A}x + \frac{C}{A})$

Complete the square for the terms involving $x$ inside the parenthesis. Add and subtract $(\frac{B}{2A})^2$:

Let $u = x + \frac{B}{2A}$. Then $du = dx$. Let $k^2 = \left|\frac{4AC - B^2}{4A^2}\right|$. The expression inside the parenthesis takes the form $u^2 + \frac{4AC-B^2}{4A^2}$. Depending on the sign of $4AC - B^2$, this will be of the form $u^2+k^2$ or $u^2-k^2$ or $k^2-u^2$ (possibly with a leading negative sign factored out with $A$). The integral then transforms into one of the standard forms (1) to (9) involving $u^2$ and $k^2$.

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Answer:

Definite Integrals as the Limit of a Sum

So far, we have dealt with indefinite integrals, which represent a family of antiderivatives. Now we transition to definite integrals. A definite integral is a numerical value that represents the signed area bounded by the graph of a function, the x-axis, and two vertical lines (the limits of integration). The formal definition of the definite integral is based on approximating this area using rectangles and taking the limit as the number of rectangles approaches infinity. This process is known as defining the definite integral as the limit of a sum, or using Riemann sums.

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Definition of Definite Integral as the Limit of a Sum (Riemann Sums)

Consider a continuous function $f(x)$ defined on a closed and bounded interval $[a, b]$. We want to find the area of the region enclosed by the curve $y=f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$. For simplicity, let's assume $f(x) \geq 0$ on $[a, b]$.

To approximate this area, we divide the interval $[a, b]$ into $n$ smaller subintervals of equal width. The width of each subinterval, denoted by $\Delta x$ or $h$, is given by:

The points that divide the interval $[a, b]$ are $x_0, x_1, x_2, \dots, x_n$, where $x_0 = a$, $x_1 = a+h$, $x_2 = a+2h$, and generally $x_r = a+rh$ for $r=0, 1, 2, \dots, n$. The last point is $x_n = a+nh = a + n\left(\frac{b-a}{n}\right) = a + (b-a) = b$.

These points divide the interval $[a, b]$ into $n$ subintervals: $[x_0, x_1], [x_1, x_2], \dots, [x_{n-1}, x_n]$. The $r$-th subinterval is $[x_{r-1}, x_r] = [a+(r-1)h, a+rh]$.

Now, we approximate the area under the curve over each subinterval by the area of a rectangle. We can choose the height of the rectangle in each subinterval as the function value at the left endpoint, right endpoint, or any point within the subinterval.

Using the right endpoint height, $f(x_r)$, for the $r$-th rectangle (over the interval $[x_{r-1}, x_r]$), the area of the $r$-th rectangle is $f(x_r) \cdot h = f(a+rh) \cdot h$. The sum of the areas of these $n$ rectangles is:

Using the left endpoint height, $f(x_{r-1})$, for the $r$-th rectangle (over the interval $[x_{r-1}, x_r]$), the area of the $r$-th rectangle is $f(x_{r-1}) \cdot h = f(a+(r-1)h) \cdot h$. The sum of the areas of these $n$ rectangles is:

These sums are called Riemann Sums. As the number of subintervals $n$ increases, the width of each subinterval $h = \frac{b-a}{n}$ decreases and approaches zero ($h \to 0$ as $n \to \infty$). The approximation of the area gets better and better.

The definite integral of $f(x)$ from $a$ to $b$ is defined as the limit of these Riemann sums as $n \to \infty$ (or as $h \to 0$):

An equivalent definition using the left endpoint sum is:

Both definitions are equivalent for continuous functions. The method using the limit of a sum provides the formal basis for the definite integral and connects it to the geometric concept of area.

Evaluation of Definite Integrals using the Limit of a Sum

To evaluate a definite integral $\int\limits_{a}^{b} f(x) dx$ using the definition as a limit of a sum, we generally follow these steps (using the formula with $r$ from 0 to $n-1$):

1. Identify the function $f(x)$, the lower limit $a$, and the upper limit $b$.

2. Calculate the width of each subinterval, $h = \frac{b-a}{n}$. Note that as $n \to \infty$, $h \to 0$.

3. Write out the terms $f(a+rh)$ for $r=0, 1, 2, \dots, n-1$.

4. Form the sum $S_n = h \sum\limits_{r=0}^{n-1} f(a+rh) = h [f(a) + f(a+h) + f(a+2h) + \dots + f(a+(n-1)h)]$.

5. Simplify the sum $S_n$. This often involves using standard formulas for sums of powers of integers or geometric series, such as:

• $\sum\limits_{r=1}^{n} r = 1+2+\dots+n = \frac{n(n+1)}{2}$
• $\sum\limits_{r=1}^{n} r^2 = 1^2+2^2+\dots+n^2 = \frac{n(n+1)(2n+1)}{6}$
• $\sum\limits_{r=1}^{n} r^3 = 1^3+2^3+\dots+n^3 = \left(\frac{n(n+1)}{2}\right)^2$
• Sum of a constant $k$ repeated $n$ times: $\sum\limits_{r=1}^{n} k = nk$

It is usually easier to work with $S_n$ expressed in terms of $n$. Remember that $h = \frac{b-a}{n}$.

6. Evaluate the limit $\lim\limits_{n \to \infty} S_n$. This is the value of the definite integral.

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Fundamental Theorems of Integral Calculus

The relationship between differentiation and integration, which appeared as the inverse operations of each other, is formally established by the Fundamental Theorem of Calculus (FTC). This theorem is a cornerstone of calculus and provides a powerful tool for evaluating definite integrals without resorting to the cumbersome method of limits of sums. The theorem is typically stated in two parts.

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First Fundamental Theorem of Calculus (Part 1)

The First Fundamental Theorem of Calculus deals with the function defined as the definite integral with a variable upper limit. It shows that the derivative of such an integral is related to the integrand.

Statement: Let $f$ be a continuous function on the closed interval $[a, b]$. Define a function $A(x)$ for $x \in [a, b]$ as the area under the curve $y=f(t)$ from $a$ to $x$:

Then the function $A(x)$ is differentiable on $(a, b)$, and its derivative is given by:

This theorem states that the rate of change of the area under the curve $y=f(t)$ with respect to the upper limit $x$ is equal to the value of the function $f$ at that upper limit $x$. This is a formal expression of the inverse relationship between differentiation and integration. The function $A(x)$ is essentially an antiderivative of $f(x)$.

Intuitive Proof Idea:

We want to find the derivative of $A(x)$, which is defined as $A'(x) = \lim\limits_{h \to 0} \frac{A(x+h) - A(x)}{h}$.

From the definition of $A(x)$:

The difference between these two definite integrals is the area under the curve $y=f(t)$ from $x$ to $x+h$.

For a small value of $h$, if $f(t)$ is continuous, its value on the small interval $[x, x+h]$ does not change significantly. The area under the curve from $x$ to $x+h$ can be approximated by the area of a rectangle with width $h$ and height approximately $f(x)$ (or $f(x+h)$, or any value between $x$ and $x+h$).

So, the difference quotient is:

Taking the limit as $h \to 0$, the approximation becomes exact because the function $f(t)$ varies less and less over the interval $[x, x+h]$ due to continuity.

Thus, $A'(x) = f(x)$. This proves that $A(x)$, the area function defined by the definite integral, is an antiderivative of $f(x)$.

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Second Fundamental Theorem of Calculus (Part 2)

The Second Fundamental Theorem of Calculus provides the practical method for evaluating definite integrals using antiderivatives.

Statement: Let $f$ be a continuous function on the closed interval $[a, b]$. If $F$ is any antiderivative of $f$ on $[a, b]$ (meaning $F'(x) = f(x)$ for all $x \in [a, b]$), then the definite integral of $f$ from $a$ to $b$ is given by the difference in the values of $F$ at the upper and lower limits:

This is the most commonly used part of the Fundamental Theorem of Calculus for evaluating definite integrals. It tells us that instead of performing the limit of a sum, we can find any antiderivative $F(x)$ of $f(x)$ and compute $F(b) - F(a)$. The constant of integration $C$ cancels out in this difference, which is why any antiderivative can be used (e.g., $(F(b)+C) - (F(a)+C) = F(b) - F(a)$).

The notation $[F(x)]_{a}^{b}$ is often used to represent $F(b) - F(a)$. So, $\int\limits_{a}^{b} f(x) dx = [F(x)]_{a}^{b}$.

Proof Idea (using the First FTC):

Let $f$ be continuous on $[a, b]$, and let $F(x)$ be any antiderivative of $f(x)$, so $F'(x) = f(x)$.

Consider the area function defined by the First FTC: $A(x) = \int\limits_{a}^{x} f(t) dt$. We know from the First FTC that $A'(x) = f(x)$.

Since both $A(x)$ and $F(x)$ are antiderivatives of the same continuous function $f(x)$, they must differ by a constant on the interval $[a, b]$. That is, there exists a constant $C$ such that:

We can find the value of $C$ by evaluating equation (B) at the lower limit $x=a$:

By the definition of $A(x)$, $A(a) = \int\limits_{a}^{a} f(t) dt$. The area under the curve from $a$ to $a$ is 0.

Substitute this value of $C$ back into equation (B):

Now, we want to evaluate the definite integral $\int\limits_{a}^{b} f(x) dx$, which by the definition of $A(x)$ is $A(b)$. Evaluate equation (C) at the upper limit $x=b$:

Since $A(b) = \int\limits_{a}^{b} f(x) dx$ (replacing the dummy variable $t$ with $x$), we have:

This completes the proof of the Second Fundamental Theorem of Calculus. It shows that the definite integral can be evaluated by finding an antiderivative.

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Evaluation of Definite Integrals by Substitution

The method of substitution (u-substitution) is not only useful for finding indefinite integrals but also for evaluating definite integrals. When applying substitution to a definite integral, it is crucial to handle the limits of integration correctly. There are two common approaches:

1. Method 1: Perform the substitution to find the indefinite integral (antiderivative) in terms of the original variable. Once the antiderivative is found, apply the original limits of integration to this result.

2. Method 2: Perform the substitution and simultaneously change the limits of integration according to the substitution. Evaluate the definite integral directly in terms of the new variable using the new limits. This method is generally more efficient as it avoids substituting back to the original variable.

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Procedure for Evaluating Definite Integrals by Substitution (Changing Limits - Method 2)

This method involves transforming the entire definite integral, including the limits, into the new variable.

To evaluate the definite integral $\int\limits_{a}^{b} f(g(x)) g'(x) dx$:

1. Make the substitution $u = g(x)$. Identify the part of the integrand that corresponds to $u$.

2. Find the differential $du$ by differentiating the substitution with respect to $x$: $du = g'(x) dx$. Identify the part of the integrand that corresponds to $du$.

3. Change the limits of integration. Since the original limits $a$ and $b$ are values of $x$, they must be converted into corresponding values of $u$ using the substitution $u = g(x)$:

• The new lower limit for $u$ is obtained by substituting the original lower limit $x=a$ into the substitution equation: $u_{lower} = g(a)$.
• The new upper limit for $u$ is obtained by substituting the original upper limit $x=b$ into the substitution equation: $u_{upper} = g(b)$.

These new limits $g(a)$ and $g(b)$ will be the limits of integration for the integral with respect to $u$.

4. Rewrite the definite integral entirely in terms of the new variable $u$ and its new limits of integration:

5. Evaluate the resulting definite integral with respect to $u$ using the Fundamental Theorem of Calculus. Find an antiderivative $F(u)$ of $f(u)$ and compute $F(g(b)) - F(g(a))$. Note that we do not need to substitute back to $x$ because the limits have been changed to correspond to $u$.

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Properties of Definite Integrals

Definite integrals possess several properties that are very useful in simplifying calculations and evaluating integrals, especially when analytical integration is difficult or impossible. These properties stem from the definition of the definite integral as the limit of a sum and the relationship between definite and indefinite integrals established by the Fundamental Theorem of Calculus.

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Basic Properties of Definite Integrals

Let $f(x)$ and $g(x)$ be integrable functions on the interval $[a, b]$.

Property P0: Change of Variable

This property states that the value of a definite integral does not depend on the choice of the variable of integration (sometimes called a "dummy variable"). It can be $x$, $t$, $u$, etc., as long as the function and the limits remain the same. This is because the definite integral represents a specific numerical value (area), which is independent of the variable used to define the function.

Property P1: Interchanging Limits

Interchanging the upper and lower limits of integration changes the sign of the definite integral.

A special case is when the upper and lower limits are the same:

This makes sense intuitively, as the area under a curve over an interval of zero width is zero.

Proof of P1 (using FTC): Let $F(x)$ be an antiderivative of $f(x)$, so $F'(x) = f(x)$. By the Second Fundamental Theorem of Calculus (FTC):

Now consider the integral with interchanged limits:

Comparing the two results:

So, $\int\limits_{b}^{a} f(x) dx = -\int\limits_{a}^{b} f(x) dx$, or equivalently $\int\limits_{a}^{b} f(x) dx = -\int\limits_{b}^{a} f(x) dx$. For the special case $\int\limits_{a}^{a} f(x) dx = F(a) - F(a) = 0$.

Property P2: Additive Property of Intervals

This property allows us to split an integral over an interval into the sum of integrals over subintervals. The point $c$ does not necessarily have to be between $a$ and $b$. For example, $\int\limits_{a}^{b} f(x) dx = \int\limits_{a}^{c} f(x) dx + \int\limits_{c}^{b} f(x) dx$ is true even if $c > b$ or $c < a$, provided $f$ is integrable on the extended interval.

Proof of P2 (using FTC): Let $F(x)$ be an antiderivative of $f(x)$.

Consider the right side of the equation:

$= F(c) - F(a) + F(b) - F(c)$

$= F(b) - F(a)$

Since both sides are equal to $F(b) - F(a)$, the property is proved.

Property P3: Linearity

where $c_1$ and $c_2$ are constants. This property states that the definite integral of a linear combination of functions is the linear combination of their definite integrals. This property is a direct consequence of the linearity properties of indefinite integrals and the FTC.

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Special Properties of Definite Integrals

These properties often help in simplifying or directly evaluating definite integrals over specific types of intervals or for functions with certain symmetries.

Property P4: Integral over $[0, a]$

This property is extremely useful for evaluating integrals where the integrand involves trigonometric functions. The property states that replacing the variable $x$ by $(a-x)$ within the integrand does not change the value of the definite integral over the interval $[0, a]$.

Proof of P4: Let the left side of the equation be $I = \int\limits_{0}^{a} f(x) dx$. In the integral $\int\limits_{0}^{a} f(a-x) dx$, let's use the substitution $t = a-x$.

Differentiating the substitution with respect to $x$: $\frac{dt}{dx} = \frac{d}{dx}(a-x) = -1$. So, the differential $dt = -dx$, which means $dx = -dt$.

Now, change the limits of integration according to the substitution $t = a-x$:

• When $x = 0$ (lower limit), $t = a - 0 = a$. The new lower limit is $a$.
• When $x = a$ (upper limit), $t = a - a = 0$. The new upper limit is $0$.

Substitute $a-x = t$ and $dx = -dt$ into the integral $\int\limits_{0}^{a} f(a-x) dx$, and use the new limits:

Move the constant $-1$ outside the integral:

Using Property P1, $\int\limits_{a}^{0} f(t) dt = -\int\limits_{0}^{a} f(t) dt$:

Using Property P0, $\int\limits_{0}^{a} f(t) dt = \int\limits_{0}^{a} f(x) dx$:

This is the original integral $I$. Thus, $\int\limits_{0}^{a} f(x) dx = \int\limits_{0}^{a} f(a-x) dx$.

Property P5: Integral over $[0, 2a]$

This is a direct application of the additive property P2. However, this property is often used in conjunction with a substitution in the second integral. In the integral $\int\limits_{a}^{2a} f(x) dx$, let $x = 2a-t$. Then $dx = -dt$. When $x=a$, $t=a$. When $x=2a$, $t=0$.

So, Property P5 is usually written as:

This property leads to a useful special case based on the symmetry of $f(x)$ about $x=a$:

Property P6: Integral over $[-a, a]$ (Even/Odd Functions)

This property applies when the interval of integration is symmetric about zero.

An even function is symmetric about the y-axis, and an odd function has rotational symmetry about the origin. The property reflects that the integral of an odd function over a symmetric interval is zero (positive and negative areas cancel out), while the integral of an even function over a symmetric interval is twice the integral over half the interval ($[0, a]$).

Proof of P6: Using Property P2, we split the integral:

Consider the first integral $\int\limits_{-a}^{0} f(x) dx$. Let's use the substitution $t = -x$. Then $dt = -dx$, so $dx = -dt$. Change the limits: when $x=-a$, $t = -(-a) = a$. When $x=0$, $t = -0 = 0$.

$= - \int\limits_{a}^{0} f(-t) dt$

Using Property P1, $- \int\limits_{a}^{0} f(-t) dt = \int\limits_{0}^{a} f(-t) dt$.

Substitute this back into equation (E):

Now, we consider the cases for even and odd functions:

If $f$ is an even function, $f(-x) = f(x)$.

If $f$ is an odd function, $f(-x) = -f(x)$.

This completes the proof of Property P6.

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