Applications of Integrals: Area Calculation

Applications of Integrals: Area Calculation

One of the most intuitive and fundamental applications of the definite integral is the calculation of area. As established by the definition of the definite integral as the limit of a Riemann sum, the integral $\int_{a}^{b} f(x) dx$ directly relates to the area of the region bounded by the curve $y=f(x)$, the x-axis, and vertical lines, under specific conditions.

Area under a Curve (Above the x-axis)

If $f(x)$ is a continuous function and $f(x) \ge 0$ for all $x$ in the closed interval $[a, b]$, then the definite integral of $f(x)$ from $a$ to $b$ gives the exact geometric area of the region situated above the x-axis and beneath the curve $y=f(x)$, between the vertical lines $x=a$ and $x=b$.

The area $A$ is given by:

$A = \int_{a}^{b} f(x) dx$

This formula directly applies when the function's graph lies entirely on or above the x-axis over the interval of integration.

Area when the Curve is Below the x-axis

If $f(x)$ is a continuous function such that $f(x) \le 0$ for all $x$ in the closed interval $[a, b]$, the definite integral $\int_{a}^{b} f(x) dx$ will yield a non-positive value (the net signed area will be negative). Since geometric area is always a non-negative quantity, to find the actual area of the region bounded by the curve $y=f(x)$, the x-axis, $x=a$, and $x=b$ when the curve is below the axis, we take the absolute value of the integral.

Area $A = \left| \int_{a}^{b} f(x) dx \right|$.

Since $f(x) \le 0$ on $[a, b]$, the value of the integral $\int_{a}^{b} f(x) dx$ is $\le 0$. The geometric area is the magnitude of this negative value, which is the negative of the integral:

$A = - \int_{a}^{b} f(x) dx$ (when $f(x) \le 0$ on $[a, b]$)

Alternatively, we can find the area by integrating the absolute value of the function: $\int_{a}^{b} |f(x)| dx$. If $f(x) \le 0$, then $|f(x)| = -f(x)$, so $\int_{a}^{b} |f(x)| dx = \int_{a}^{b} (-f(x)) dx = - \int_{a}^{b} f(x) dx$.

Area when the Curve Crosses the x-axis

If the continuous function $f(x)$ takes on both positive and negative values over the interval $[a, b]$, the definite integral $\int_{a}^{b} f(x) dx$ gives the net signed area (positive area minus negative area). To find the total geometric area of the regions bounded by the curve and the x-axis, we must treat the areas above and below the axis as positive quantities and sum them.

To do this, we first find the points where the function crosses the x-axis within the interval $(a, b)$ by solving $f(x) = 0$. Let these roots be $c_1, c_2, \dots, c_k$ such that $a < c_1 < c_2 < \dots < c_k < b$. These points divide the interval $[a, b]$ into subintervals where the function's sign is constant.

The total geometric area is the sum of the absolute values of the definite integrals over each subinterval:

Total Geometric Area $= \int_{a}^{b} |f(x)| dx = \int_{a}^{c_1} |f(x)| dx + \int_{c_1}^{c_2} |f(x)| dx + \dots + \int_{c_k}^{b} |f(x)| dx$

On each subinterval, replace $|f(x)|$ with either $f(x)$ (if $f(x) \ge 0$) or $-f(x)$ (if $f(x) \le 0$) before evaluating the integral.

For example, if $f(x)$ is positive on $[a, c]$ and negative on $[c, b]$, the total area is:

Total Area $= \int_{a}^{c} f(x) dx + \int_{c}^{b} (-f(x)) dx = \int_{a}^{c} f(x) dx - \int_{c}^{b} f(x) dx$

This is equivalent to adding the absolute values of the individual integrals: $|\int_{a}^{c} f(x) dx| + |\int_{c}^{b} f(x) dx|$, since $\int_{a}^{c} f(x) dx \ge 0$ and $\int_{c}^{b} f(x) dx \le 0$.

Area with Respect to the y-axis

Sometimes, it is easier to define a region by bounding it with curves expressed as functions of $y$ ($x=g(y)$) and horizontal lines ($y=c$ and $y=d$). In this case, the area can be calculated by integrating with respect to $y$.

If $x=g(y)$ is a continuous function such that $g(y) \ge 0$ for all $y$ in the closed interval $[c, d]$, then the area $A$ of the region bounded by the curve $x=g(y)$, the y-axis ($x=0$), and the horizontal lines $y=c$ and $y=d$ is given by:

$A = \int_{c}^{d} g(y) dy$

Similar to the x-axis case, if $g(y) \le 0$ on $[c, d]$, the area is $- \int_{c}^{d} g(y) dy$. If $g(y)$ changes sign, you find the roots of $g(y)=0$ and split the integral based on $|g(y)|$ with respect to $y$.

Answer:

Area Between Two Curves

The definite integral can also be used to calculate the area of a region bounded between the graphs of two functions. This is a natural extension of finding the area between a curve and the x-axis (which is just the area between the curve and the line $y=0$).

Area Between $y=f(x)$ and $y=g(x)$ (Integration with respect to $x$)

Let $f(x)$ and $g(x)$ be two continuous functions over the closed interval $[a, b]$. If $f(x) \ge g(x)$ for all $x$ in $[a, b]$ (meaning the graph of $f$ is always above or touching the graph of $g$ over this interval), then the area $A$ of the region bounded by the curves $y=f(x)$, $y=g(x)$, and the vertical lines $x=a$ and $x=b$ is given by the integral of the difference between the upper function and the lower function:

$A = \int_{a}^{b} [f(x) - g(x)] dx$

This formula works even if one or both functions are negative or cross the x-axis within $[a, b]$, as long as $f(x) \ge g(x)$ throughout the interval. The difference $f(x) - g(x)$ represents the vertical distance between the two curves at each $x$. Integrating this distance over the interval $[a, b]$ sums up the areas of infinitesimally thin vertical strips, giving the total area between the curves.

Finding the Limits of Integration: If the vertical boundary lines $x=a$ and $x=b$ are not explicitly given, the region might be bounded entirely by the curves themselves. In such cases, the limits of integration $a$ and $b$ are the x-coordinates of the points where the two curves intersect. To find these, set $f(x) = g(x)$ and solve for $x$. The smallest solution is typically the lower limit $a$, and the largest solution is the upper limit $b$, assuming there are only two intersection points bounding the region.

Multiple Intersection Points: If the curves intersect at more than two points over the region of interest, the roles of the upper and lower curves might switch. You need to find all intersection points in the relevant interval, split the area calculation into multiple integrals over the sub-intervals defined by these points, and in each sub-interval, ensure you are integrating the upper curve minus the lower curve.

Area Between $x=h(y)$ and $x=k(y)$ (Integration with respect to $y$)

For some regions, it is easier to describe the bounding curves as functions of $y$, i.e., $x=h(y)$ and $x=k(y)$. If the region is bounded by these curves and horizontal lines $y=c$ and $y=d$, and if $h(y) \ge k(y)$ for all $y$ in the interval $[c, d]$ (meaning the graph of $h$ is always to the right of or touching the graph of $k$ over this interval), the area $A$ is given by integrating with respect to $y$:

$A = \int_{c}^{d} [h(y) - k(y)] dy$

This formula integrates the horizontal distance between the two curves ($h(y) - k(y)$, the "right curve" minus the "left curve") over the interval $[c, d]$ on the y-axis. The limits $c$ and $d$ are often the y-coordinates of the points where the curves intersect, found by setting $h(y)=k(y)$ and solving for $y$.

Answer:

The region is enclosed by two curves: $y = x^2$ and $y = 2x - x^2$. The boundaries of the region are the curves themselves. We need to find the points of intersection to determine the limits of integration.

Step 1: Find the Intersection Points (Limits of Integration).

To find where the curves intersect, set their $y$-values equal to each other:

"$x^2 = 2x - x^2$"

[Set equations equal]

Move all terms to one side to solve for $x$:

"$x^2 + x^2 - 2x = 0$"

"$2x^2 - 2x = 0$"

Factor out the common term $2x$:

"$2x(x - 1) = 0$"

[Factor]

This gives two solutions for $x$:

"$2x = 0 \implies x = 0$"

"$x - 1 = 0 \implies x = 1$"

The curves intersect at $x=0$ and $x=1$. These will serve as our limits of integration, so $a=0$ and $b=1$.

Step 2: Determine Which Curve is Upper and Which is Lower in $[0, 1]$.

We need to know which function is $f(x)$ (the upper curve) and which is $g(x)$ (the lower curve) on the interval $(0, 1)$. We can pick a test value within this interval, say $x = 0.5$, and evaluate both functions:

For $y = x^2$: At $x=0.5$, $y = (0.5)^2 = 0.25$.

For $y = 2x - x^2$: At $x=0.5$, $y = 2(0.5) - (0.5)^2 = 1 - 0.25 = 0.75$.

Since $0.75 > 0.25$, the curve $y = 2x - x^2$ is above $y = x^2$ in the interval $(0, 1)$. So, $f(x) = 2x - x^2$ is the upper curve and $g(x) = x^2$ is the lower curve for $x \in [0, 1]$.

Step 3: Set Up the Integral for the Area.

Using the formula for the area between two curves $A = \int_{a}^{b} [f(x) - g(x)] dx$, with $a=0$, $b=1$, $f(x) = 2x - x^2$, and $g(x) = x^2$:

"$A = \int_{0}^{1} [(2x - x^2) - (x^2)] dx$"

[Integral of upper minus lower curve]

Simplify the integrand:

"$A = \int_{0}^{1} (2x - x^2 - x^2) dx = \int_{0}^{1} (2x - 2x^2) dx$"

Step 4: Evaluate the Integral.

Evaluate the definite integral using the Second Fundamental Theorem of Calculus.

An antiderivative of $(2x - 2x^2)$ is found using linearity and the Power Rule:

$\int (2x - 2x^2) dx = 2 \int x dx - 2 \int x^2 dx = 2 \left(\frac{x^2}{2}\right) - 2 \left(\frac{x^3}{3}\right) = x^2 - \frac{2}{3}x^3$.

Let $F(x) = x^2 - \frac{2}{3}x^3$.

"$A = \left[ x^2 - \frac{2}{3}x^3 \right]_{0}^{1}$"

[Evaluate using FTC2]

Evaluate at the upper limit ($x=1$) and subtract the evaluation at the lower limit ($x=0$):

"$= \left( (1)^2 - \frac{2}{3}(1)^3 \right) - \left( (0)^2 - \frac{2}{3}(0)^3 \right)$"

[Evaluate at limits and subtract]

$= \left( 1 - \frac{2}{3}(1) \right) - (0 - \frac{2}{3}(0))$"

$= \left( 1 - \frac{2}{3} \right) - (0)$"

Calculate the difference in the first bracket:

"$1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}$"

"$A = \frac{1}{3}$"

The area of the enclosed region is $\frac{1}{3}$ square units.

Application of Integration in finding Areas (Applied Maths)

From an applied mathematics, science, or engineering perspective, the calculation of area using definite integrals is far more than a theoretical exercise. It is a fundamental tool for quantifying spatial extent, accumulation, or physical quantities represented by areas under or between curves in various contexts. Integration for area calculation is a direct application of the definition of the definite integral as the limit of a sum.

Core Process in Applied Area Problems

Solving applied problems that involve finding areas using integration follows a general process, similar to the steps for evaluating definite integrals but with an added emphasis on modeling and interpretation:

1. Define and Understand the Region: Clearly identify the region whose area needs to be calculated. This region is typically bounded by the graphs of one or more functions and possibly horizontal or vertical lines. Translate the problem description into mathematical equations defining the boundaries.
2. Visualize the Region and Set Up the Integral:
   • Sketching the Graph: Drawing a clear diagram of the bounding curves is almost always essential. This visualization helps in:
     • Determining the limits of integration (the x or y values where the region starts and ends). These might be given directly by lines ($x=a, x=b$ or $y=c, y=d$) or they might be the intersection points of the bounding curves (found by setting the function equations equal to each other).
     • Identifying which function is the "upper" curve and which is the "lower" curve (if integrating with respect to $x$), or which is the "right" curve and which is the "left" curve (if integrating with respect to $y$).
     • Deciding whether it is more convenient to integrate with respect to $x$ ($dx$) or $y$ ($dy$). Sometimes one choice avoids splitting the integral into multiple parts or simplifies the functions involved.
   • Setting Up the Definite Integral: Based on the decision to integrate with respect to $x$ or $y$, set up the definite integral.
     • If integrating with respect to $x$: $A = \int_{a}^{b} [\text{Upper Function}(x) - \text{Lower Function}(x)] dx$.
     • If integrating with respect to $y$: $A = \int_{c}^{d} [\text{Right Function}(y) - \text{Left Function}(y)] dy$.
3. Evaluate the Definite Integral: Use the techniques of integral calculus (Fundamental Theorem of Calculus, substitution, integration by parts, standard formulas, etc.) to evaluate the definite integral and find the numerical value of the area.
4. Interpret the Result: State the final answer clearly, in the context of the original problem, including appropriate units (e.g., square meters, square inches).

This systematic approach ensures that the integral is set up correctly to represent the desired area and that the result is interpreted meaningfully in the applied context.

Significance and Examples in Applied Fields

The ability to calculate areas using integration has widespread significance across various disciplines:

• Engineering and Physics:
  • Work Done by a Variable Force: If a force $F(x)$ varies with displacement $x$, the work done in moving an object from $a$ to $b$ is the area under the Force vs. Displacement curve: $W = \int_{a}^{b} F(x) dx$.
  • Impulse: If a force $F(t)$ varies with time $t$, the impulse delivered from $t_1$ to $t_2$ is the area under the Force vs. Time curve: $Impulse = \int_{t_1}^{t_2} F(t) dt$. Impulse is equal to the change in momentum.
  • Fluid Flow: Calculating the total flow rate through a pipe by integrating the fluid velocity profile (velocity as a function of distance from the center).
  • Charge from Current: Finding the total electric charge that flows through a point in a circuit over a time interval by integrating the electric current $I(t)$ (rate of charge flow) with respect to time: $Q = \int_{t_1}^{t_2} I(t) dt$.
  • Cross-sectional Areas: Calculating the area of cross-sections of various objects (beams, pipes, aircraft wings) which is crucial for stress analysis, material properties, and fluid dynamics.
• Economics:
  • Consumer Surplus and Producer Surplus: These concepts represent the total benefit to consumers and producers in a market and are calculated as areas between the demand/supply curves and the market price line.
  • Total Cost/Revenue/Profit from Marginal Functions: While not strictly "area" under a curve in the geometric sense, integrating marginal cost, revenue, or profit functions (rates of change) gives the total cost, revenue, or profit function (accumulation), which is an application of the fundamental relationship between integral and derivative. For example, Total Cost = $\int$ (Marginal Cost) $dx$.
• Biology and Environmental Science:
  • Population Change: Calculating the total change in population size over a time interval by integrating the population growth rate.
  • Accumulation of Substances: Finding the total amount of a substance accumulated over time or space by integrating its rate of accumulation or concentration.
• Statistics and Probability:
  • Probability: For a continuous random variable with probability density function $f(x)$, the probability that the variable falls within an interval $[a, b]$ is given by the area under the curve $y=f(x)$ from $a$ to $b$: $P(a \le X \le b) = \int_{a}^{b} f(x) dx$. The total area under a probability density function over its entire domain is always 1.

The representation and calculation of area using definite integrals provide a fundamental tool for quantifying cumulative effects and total quantities from rates of change or spatial distributions in a wide array of scientific, engineering, and economic problems.