Chapter 8 Application Of Integrals

Given:

The equation of the circle is $x^2 + y^2 = a^2$.

To Find:

The area enclosed by the circle.

Solution:

The given equation $x^2 + y^2 = a^2$ represents a circle centered at the origin $(0,0)$ with radius $a$.

To find the area enclosed by the circle, we can use integration. Due to the symmetry of the circle, we can calculate the area of one quarter of the circle (e.g., in the first quadrant) and multiply the result by 4.

In the first quadrant, $x \geq 0$ and $y \geq 0$. From the equation $x^2 + y^2 = a^2$, we solve for $y$: $y^2 = a^2 - x^2$. Since $y \geq 0$, we have $y = \sqrt{a^2 - x^2}$.

The area of the portion of the circle in the first quadrant is given by the definite integral of $y$ with respect to $x$ from $x=0$ to $x=a$.

Area (A) $= 4 \int_{0}^{a} y \, dx$

Substitute $y = \sqrt{a^2 - x^2}$:

A $= 4 \int_{0}^{a} \sqrt{a^2 - x^2} \, dx$

We use the standard integral formula: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$.

Now, we evaluate the definite integral from $0$ to $a$:

A $= 4 \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a}$

Applying the upper limit ($x=a$):

$ \left[ \frac{a}{2}\sqrt{a^2 - a^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{a}{a}\right) \right] = \left[ \frac{a}{2}\sqrt{0} + \frac{a^2}{2}\sin^{-1}(1) \right] $

$= \left[ 0 + \frac{a^2}{2} \cdot \frac{\pi}{2} \right] = \frac{\pi a^2}{4} $

Applying the lower limit ($x=0$):

$ \left[ \frac{0}{2}\sqrt{a^2 - 0^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{0}{a}\right) \right] = \left[ 0 + \frac{a^2}{2}\sin^{-1}(0) \right] $

$= \left[ 0 + \frac{a^2}{2} \cdot 0 \right] = 0 $

Subtracting the lower limit value from the upper limit value:

$ \left[ \frac{\pi a^2}{4} - 0 \right] = \frac{\pi a^2}{4} $

Finally, multiply this result by 4 to get the total area of the circle:

A $= 4 \times \frac{\pi a^2}{4}$

A $= \pi a^2$

Conclusion:

The area enclosed by the circle $x^2 + y^2 = a^2$ is $\pi a^2$ square units.

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

To Find:

The area enclosed by the ellipse.

Solution:

The given equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ represents an ellipse centered at the origin $(0,0)$ with semi-major and semi-minor axes along the x and y axes, respectively. The x-intercepts are $(\pm a, 0)$ and the y-intercepts are $(0, \pm b)$.

To find the area enclosed by the ellipse, we can use integration. Due to the symmetry of the ellipse with respect to both axes, we can calculate the area of one quarter of the ellipse (e.g., in the first quadrant) and multiply the result by 4.

In the first quadrant, $x \geq 0$ and $y \geq 0$. From the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we solve for $y$: $\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2}$

$\frac{y^2}{b^2} = \frac{a^2 - x^2}{a^2}$

$y^2 = \frac{b^2}{a^2}(a^2 - x^2)$

Since $y \geq 0$ in the first quadrant, we have $y = \sqrt{\frac{b^2}{a^2}(a^2 - x^2)} = \frac{b}{a}\sqrt{a^2 - x^2}$.

The area of the portion of the ellipse in the first quadrant is given by the definite integral of $y$ with respect to $x$ from $x=0$ to $x=a$.

Area (A) $= 4 \int_{0}^{a} y \, dx$

Substitute $y = \frac{b}{a}\sqrt{a^2 - x^2}$:

A $= 4 \int_{0}^{a} \frac{b}{a}\sqrt{a^2 - x^2} \, dx$

A $= \frac{4b}{a} \int_{0}^{a} \sqrt{a^2 - x^2} \, dx$

We use the standard integral formula: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$.

Now, we evaluate the definite integral from $0$ to $a$:

A $= \frac{4b}{a} \left[ \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_{0}^{a}$

Applying the upper limit ($x=a$):

$ \left[ \frac{a}{2}\sqrt{a^2 - a^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{a}{a}\right) \right] = \left[ \frac{a}{2}\sqrt{0} + \frac{a^2}{2}\sin^{-1}(1) \right] $

$= \left[ 0 + \frac{a^2}{2} \cdot \frac{\pi}{2} \right] = \frac{\pi a^2}{4} $

Applying the lower limit ($x=0$):

$ \left[ \frac{0}{2}\sqrt{a^2 - 0^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{0}{a}\right) \right] = \left[ 0 + \frac{a^2}{2}\sin^{-1}(0) \right] $

$= \left[ 0 + \frac{a^2}{2} \cdot 0 \right] = 0 $

Subtracting the lower limit value from the upper limit value:

$ \left[ \frac{\pi a^2}{4} - 0 \right] = \frac{\pi a^2}{4} $

Substituting this value back into the expression for A:

A $= \frac{4b}{a} \times \frac{\pi a^2}{4}$

A $= \pi a b$

Conclusion:

The area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi ab$ square units.

Given:

The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

To Find:

The area of the region bounded by the ellipse.

Solution:

The given equation is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

Comparing this with the standard equation of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we have $a^2 = 16$ and $b^2 = 9$.

So, $a = \sqrt{16} = 4$ and $b = \sqrt{9} = 3$.

The ellipse is centered at the origin $(0,0)$ with semi-major axis $a=4$ along the x-axis and semi-minor axis $b=3$ along the y-axis.

To find the area enclosed by the ellipse, we use integration. Due to symmetry, we can find the area of the part of the ellipse in the first quadrant ($x \geq 0, y \geq 0$) and multiply it by 4.

From the equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$, we solve for $y$ in the first quadrant:

$\frac{y^2}{9} = 1 - \frac{x^2}{16}$

$\frac{y^2}{9} = \frac{16 - x^2}{16}$

$y^2 = \frac{9}{16}(16 - x^2)$

Since $y \geq 0$, $y = \sqrt{\frac{9}{16}(16 - x^2)} = \frac{3}{4}\sqrt{16 - x^2}$.

The area of the ellipse in the first quadrant is given by the integral $\int_{0}^{4} y \, dx$. The x-values range from 0 to $a=4$ in the first quadrant.

Area (A) $= 4 \int_{0}^{4} y \, dx$

Substitute $y = \frac{3}{4}\sqrt{16 - x^2}$:

A $= 4 \int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2} \, dx$

A $= 3 \int_{0}^{4} \sqrt{4^2 - x^2} \, dx$

Using the standard integral formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$, with $a=4$:

A $= 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4}$

A $= 3 \left[ \frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4}$

Evaluate at the upper limit ($x=4$):

$ \left( \frac{4}{2}\sqrt{16 - 4^2} + 8\sin^{-1}\left(\frac{4}{4}\right) \right) = \left( 2\sqrt{16 - 16} + 8\sin^{-1}(1) \right) $

$= \left( 2\sqrt{0} + 8 \cdot \frac{\pi}{2} \right) = (0 + 4\pi) = 4\pi $

Evaluate at the lower limit ($x=0$):

$ \left( \frac{0}{2}\sqrt{16 - 0^2} + 8\sin^{-1}\left(\frac{0}{4}\right) \right) = \left( 0\sqrt{16} + 8\sin^{-1}(0) \right) $

$= (0 + 8 \cdot 0) = 0 $

Subtract the lower limit value from the upper limit value:

A $= 3 (4\pi - 0) = 12\pi$

Conclusion:

The area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is $12\pi$ square units.

Alternate Solution (Using Formula):

The given equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

This is in the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 16$ and $b^2 = 9$.

Therefore, $a = 4$ and $b = 3$.

The area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is given by the formula $\pi ab$.

Area $= \pi \times a \times b$

Area $= \pi \times 4 \times 3$

Area $= 12\pi$

Thus, the area of the region bounded by the ellipse is $12\pi$ square units.

Given:

The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

To Find:

The area of the region bounded by the ellipse.

Solution:

The given equation is $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

This equation is in the standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 4$ and $b^2 = 9$.

Thus, $a = \sqrt{4} = 2$ and $b = \sqrt{9} = 3$.

The ellipse is centered at the origin $(0,0)$. Since $b > a$, the semi-major axis is along the y-axis and the semi-minor axis is along the x-axis.

To find the area enclosed by the ellipse, we can use integration. Due to the symmetry of the ellipse with respect to both the x and y axes, we can calculate the area of the portion of the ellipse in the first quadrant ($x \geq 0, y \geq 0$) and multiply it by 4.

From the equation $\frac{x^2}{4} + \frac{y^2}{9} = 1$, we solve for $y$ in terms of $x$ for the first quadrant:

$\frac{y^2}{9} = 1 - \frac{x^2}{4}$

$\frac{y^2}{9} = \frac{4 - x^2}{4}$

$y^2 = \frac{9}{4}(4 - x^2)$

Since $y \geq 0$ in the first quadrant, $y = \sqrt{\frac{9}{4}(4 - x^2)} = \frac{3}{2}\sqrt{4 - x^2}$.

In the first quadrant, the x-values range from 0 to $a=2$. The area of the ellipse in the first quadrant is given by the definite integral $\int_{0}^{2} y \, dx$.

Total Area (A) $= 4 \times \int_{0}^{2} y \, dx$

Substitute the expression for $y$:

A $= 4 \int_{0}^{2} \frac{3}{2}\sqrt{4 - x^2} \, dx$

A $= 6 \int_{0}^{2} \sqrt{2^2 - x^2} \, dx$

Using the standard integral formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$, with $a=2$:

A $= 6 \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

A $= 6 \left[ \frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

Now, we evaluate the expression at the limits of integration.

At the upper limit $x=2$:

$ \left( \frac{2}{2}\sqrt{4 - 2^2} + 2\sin^{-1}\left(\frac{2}{2}\right) \right) = \left( 1\sqrt{4 - 4} + 2\sin^{-1}(1) \right) $

$= \left( 1\sqrt{0} + 2 \cdot \frac{\pi}{2} \right) = (0 + \pi) = \pi $

At the lower limit $x=0$:

$ \left( \frac{0}{2}\sqrt{4 - 0^2} + 2\sin^{-1}\left(\frac{0}{2}\right) \right) = \left( 0\sqrt{4} + 2\sin^{-1}(0) \right) $

$= (0 + 2 \cdot 0) = 0 $

Subtracting the lower limit value from the upper limit value:

$ \pi - 0 = \pi $

So, the definite integral evaluates to $\pi$.

The total area A is $6$ times this value:

A $= 6 \times \pi = 6\pi$

Conclusion:

The area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$ is $6\pi$ square units.

Alternate Solution (Using Formula):

The given equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

This is in the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2 = 4$ and $b^2 = 9$.

Therefore, $a = \sqrt{4} = 2$ and $b = \sqrt{9} = 3$.

The area of an ellipse with semi-major axis $b$ and semi-minor axis $a$ (or vice-versa) is given by the formula $\pi ab$.

Area $= \pi \times a \times b$

Area $= \pi \times 2 \times 3$

Area $= 6\pi$

Thus, the area of the region bounded by the ellipse is $6\pi$ square units.

Given:

Region bounded by the circle $x^2 + y^2 = 4$, the line $x=0$ (y-axis), and the line $x=2$, in the first quadrant.

To Find:

The area of the given region.

Solution:

The given equation of the circle is $x^2 + y^2 = 4$. This is a circle centered at the origin $(0,0)$ with radius $r = \sqrt{4} = 2$.

We need to find the area of the region in the first quadrant ($x \geq 0, y \geq 0$) bounded by this circle, the y-axis ($x=0$), and the vertical line $x=2$.

In the first quadrant, the equation of the circle is $y = \sqrt{4 - x^2}$.

The region is bounded by $x=0$ and $x=2$. For a circle of radius 2, the portion in the first quadrant extends from $x=0$ to $x=2$. Thus, the given lines $x=0$ and $x=2$ encompass the entire width of the quarter circle in the first quadrant.

The area of this region can be found by integrating the function $y = \sqrt{4 - x^2}$ with respect to $x$ from $x=0$ to $x=2$.

Area (A) $= \int_{0}^{2} \sqrt{4 - x^2} \, dx$

This integral is of the form $\int \sqrt{a^2 - x^2} \, dx$, where $a = 2$. We use the standard formula: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$.

Applying the formula with $a=2$:

A $= \left[ \frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

A $= \left[ \frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{0}^{2}$

Now, evaluate the expression at the upper limit ($x=2$):

$ \text{Value at upper limit} = \frac{2}{2}\sqrt{4 - 2^2} + 2\sin^{-1}\left(\frac{2}{2}\right) $

$= 1\sqrt{4 - 4} + 2\sin^{-1}(1) $

$= 1\sqrt{0} + 2 \cdot \frac{\pi}{2} $

$= 0 + \pi = \pi $

Evaluate the expression at the lower limit ($x=0$):

$ \text{Value at lower limit} = \frac{0}{2}\sqrt{4 - 0^2} + 2\sin^{-1}\left(\frac{0}{2}\right) $

$= 0\sqrt{4} + 2\sin^{-1}(0) $

$= 0 + 2 \cdot 0 = 0 $

Subtract the value at the lower limit from the value at the upper limit:

A $= \pi - 0 = \pi$

Thus, the area of the specified region is $\pi$ square units.

Conclusion:

The area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is $\pi$.

Comparing with the given options:

(A) $\pi$

(B) $\frac{\pi}{2}$

(C) $\frac{\pi}{3}$

(D) $\frac{\pi}{4}$

The correct answer is (A).

Given:

Region bounded by the curve $y^2 = 4x$, the y-axis ($x=0$), and the line $y=3$.

To Find:

The area of the region bounded by the given curves.

Solution:

The given curve is $y^2 = 4x$. This is a parabola opening towards the positive x-axis, with its vertex at the origin $(0,0)$.

The region is bounded by the parabola $y^2 = 4x$, the y-axis ($x=0$), and the horizontal line $y=3$.

To find the area of this region, it is convenient to integrate with respect to $y$. We express $x$ in terms of $y$ from the equation of the parabola:

$x = \frac{y^2}{4}$

The region is bounded on the left by the y-axis ($x=0$) and on the right by the curve $x = \frac{y^2}{4}$. The y-values range from $y=0$ (the origin) up to the line $y=3$.

The area (A) of the region is given by the integral of $x$ with respect to $y$ from $y=0$ to $y=3$:

A $= \int_{0}^{3} x \, dy$

Substitute $x = \frac{y^2}{4}$:

A $= \int_{0}^{3} \frac{y^2}{4} \, dy$

Now, we evaluate the definite integral:

A $= \frac{1}{4} \int_{0}^{3} y^2 \, dy$

A $= \frac{1}{4} \left[ \frac{y^{2+1}}{2+1} \right]_{0}^{3}$

A $= \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3}$

Now, evaluate the expression at the limits of integration:

A $= \frac{1}{4} \left( \frac{3^3}{3} - \frac{0^3}{3} \right)$

A $= \frac{1}{4} \left( \frac{27}{3} - 0 \right)$

A $= \frac{1}{4} (9)$

A $= \frac{9}{4}$

The area of the region is $\frac{9}{4}$ square units.

Conclusion:

The area of the region bounded by the curve $y^2 = 4x$, y-axis and the line $y = 3$ is $\frac{9}{4}$.

Comparing with the given options:

(A) 2

(B) $\frac{9}{4}$

(C) $\frac{9}{3} = 3$

(D) $\frac{9}{2}$

The correct answer is (B).

Given:

The region bounded by the line $y = 3x + 2$, the x-axis ($y=0$), and the ordinates $x = –1$ and $x = 1$.

To Find:

The area of the specified region.

Solution:

The given line is $y = 3x + 2$. The region is bounded by this line, the x-axis ($y=0$), and the vertical lines $x=-1$ and $x=1$.

To find the area of the region between a curve $y=f(x)$ and the x-axis from $x=a$ to $x=b$, we calculate $\int_{a}^{b} f(x) \, dx$. However, if the curve goes below the x-axis in the interval $[a, b]$, the integral will give a negative value for the area below the axis. Since area is a positive quantity, we need to take the absolute value of the integral over any interval where the function is negative.

First, let's find where the line $y = 3x + 2$ crosses the x-axis by setting $y=0$:

$3x + 2 = 0$

$3x = -2$

$x = -\frac{2}{3}$

The x-intercept is at $x = -\frac{2}{3}$. This point lies within the interval $[-1, 1]$. Specifically, it is between $x=-1$ and $x=1$.

For $x \in [-1, -\frac{2}{3}]$, the line $y = 3x + 2$ is below the x-axis (since $3x+2 < 0$ for $x < -\frac{2}{3}$).

For $x \in [-\frac{2}{3}, 1]$, the line $y = 3x + 2$ is above the x-axis (since $3x+2 > 0$ for $x > -\frac{2}{3}$).

Therefore, the total area is the sum of the absolute values of the integrals over the two sub-intervals $[-1, -\frac{2}{3}]$ and $[-\frac{2}{3}, 1]$:

Area (A) $= \int_{-1}^{1} |3x + 2| \, dx$

A $= \int_{-1}^{-2/3} |3x + 2| \, dx + \int_{-2/3}^{1} |3x + 2| \, dx$

Since $3x+2 \leq 0$ on $[-1, -2/3]$ and $3x+2 \geq 0$ on $[-2/3, 1]$:

A $= \int_{-1}^{-2/3} -(3x + 2) \, dx + \int_{-2/3}^{1} (3x + 2) \, dx$

A $= -\int_{-1}^{-2/3} (3x + 2) \, dx + \int_{-2/3}^{1} (3x + 2) \, dx$

First, let's find the indefinite integral of $3x+2$:

$\int (3x + 2) \, dx = \frac{3x^2}{2} + 2x + C$

Now, evaluate the definite integrals:

$ \int_{-1}^{-2/3} (3x + 2) \, dx = \left[ \frac{3x^2}{2} + 2x \right]_{-1}^{-2/3} $

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

$ = \left( \frac{3(\frac{4}{9})}{2} - \frac{4}{3} \right) - \left( \frac{3}{2} - 2 \right) $

$ = \left( \frac{\frac{4}{3}}{2} - \frac{4}{3} \right) - \left( \frac{3 - 4}{2} \right) $

$ = \left( \frac{4}{6} - \frac{4}{3} \right) - \left( -\frac{1}{2} \right) $

$ = \left( \frac{2}{3} - \frac{4}{3} \right) + \frac{1}{2} $

$ = -\frac{2}{3} + \frac{1}{2} = \frac{-4 + 3}{6} = -\frac{1}{6} $

Next, evaluate the second definite integral:

$ \int_{-2/3}^{1} (3x + 2) \, dx = \left[ \frac{3x^2}{2} + 2x \right]_{-2/3}^{1} $

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

$ = \left( \frac{3}{2} + 2 \right) - \left( \frac{3(\frac{4}{9})}{2} - \frac{4}{3} \right) $

$ = \left( \frac{3 + 4}{2} \right) - \left( \frac{\frac{4}{3}}{2} - \frac{4}{3} \right) $

$ = \frac{7}{2} - \left( \frac{2}{3} - \frac{4}{3} \right) $

$ = \frac{7}{2} - \left( -\frac{2}{3} \right) = \frac{7}{2} + \frac{2}{3} $

$ = \frac{21 + 4}{6} = \frac{25}{6} $

Now, substitute these values back into the area formula:

A $= -(-\frac{1}{6}) + (\frac{25}{6})$

A $= \frac{1}{6} + \frac{25}{6} = \frac{1 + 25}{6} = \frac{26}{6} = \frac{13}{3}$

Conclusion:

The area of the region bounded by the line $y = 3x + 2$, the x-axis, and the ordinates $x = –1$ and $x = 1$ is $\frac{13}{3}$ square units.

Given:

The curve $y = \cos x$ and the interval $x \in [0, 2\pi]$. The other boundary is the x-axis ($y=0$).

To Find:

The area bounded by the curve, the x-axis, and the given limits.

Solution:

The area bounded by the curve $y = f(x)$ and the x-axis from $x=a$ to $x=b$ is given by $\int_{a}^{b} |f(x)| \, dx$. We need to find the intervals where $y = \cos x$ is positive and where it is negative in the interval $[0, 2\pi]$.

The cosine function is zero when $x = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$. Within the interval $[0, 2\pi]$, the relevant points are $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.

These points divide the interval $[0, 2\pi]$ into three sub-intervals:

1. $[0, \frac{\pi}{2}]$: $\cos x \geq 0$ in this interval.

2. $[\frac{\pi}{2}, \frac{3\pi}{2}]$: $\cos x \leq 0$ in this interval.

3. $[\frac{3\pi}{2}, 2\pi]$: $\cos x \geq 0$ in this interval.

The total area (A) is the sum of the areas of the regions in these sub-intervals, taking the absolute value of the integral where $\cos x$ is negative.

A $= \int_{0}^{2\pi} |\cos x| \, dx$

A $= \int_{0}^{\pi/2} |\cos x| \, dx + \int_{\pi/2}^{3\pi/2} |\cos x| \, dx + \int_{3\pi/2}^{2\pi} |\cos x| \, dx$

A $= \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{3\pi/2} (-\cos x) \, dx + \int_{3\pi/2}^{2\pi} \cos x \, dx$

A $= \int_{0}^{\pi/2} \cos x \, dx - \int_{\pi/2}^{3\pi/2} \cos x \, dx + \int_{3\pi/2}^{2\pi} \cos x \, dx$

Now, we evaluate each integral. The indefinite integral of $\cos x$ is $\sin x$.

$\int_{0}^{\pi/2} \cos x \, dx = [\sin x]_{0}^{\pi/2} = \sin(\frac{\pi}{2}) - \sin(0) = 1 - 0 = 1$

$\int_{\pi/2}^{3\pi/2} \cos x \, dx = [\sin x]_{\pi/2}^{3\pi/2} = \sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2}) = -1 - 1 = -2$

$\int_{3\pi/2}^{2\pi} \cos x \, dx = [\sin x]_{3\pi/2}^{2\pi} = \sin(2\pi) - \sin(\frac{3\pi}{2}) = 0 - (-1) = 1$

Substitute these values back into the expression for the total area:

A $= (1) - (-2) + (1)$

A $= 1 + 2 + 1 = 4$

Conclusion:

The area bounded by the curve $y = \cos x$ between $x = 0$ and $x = 2\pi$ is $4$ square units.

Part (i)

Given:

The curve $y = x^2$, the lines $x = 1$, $x = 2$, and the x-axis.

To Find:

The area under the curve $y = x^2$ between $x = 1$ and $x = 2$, and the x-axis.

Solution:

The region whose area is to be found is bounded by the curve $y = x^2$, the vertical lines $x=1$ and $x=2$, and the x-axis ($y=0$).

Since the curve $y = x^2$ is above the x-axis for $x \in [1, 2]$, the area can be calculated directly by the definite integral of $y$ with respect to $x$ from $x=1$ to $x=2$.

Area (A) $= \int_{1}^{2} y \, dx$

Substitute $y = x^2$:

A $= \int_{1}^{2} x^2 \, dx$

Using the power rule of integration $\int x^n \, dx = \frac{x^{n+1}}{n+1}$:

A $= \left[ \frac{x^{2+1}}{2+1} \right]_{1}^{2}$

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

Evaluate the expression at the limits of integration:

A $= \frac{2^3}{3} - \frac{1^3}{3}$

A $= \frac{8}{3} - \frac{1}{3}$

A $= \frac{8 - 1}{3}$

A $= \frac{7}{3}$

Conclusion:

The area under the given curve and lines is $\frac{7}{3}$ square units.

Part (ii)

Given:

The curve $y = x^4$, the lines $x = 1$, $x = 5$, and the x-axis.

To Find:

The area under the curve $y = x^4$ between $x = 1$ and $x = 5$, and the x-axis.

Solution:

The region whose area is to be found is bounded by the curve $y = x^4$, the vertical lines $x=1$ and $x=5$, and the x-axis ($y=0$).

Since the curve $y = x^4$ is above the x-axis for $x \in [1, 5]$, the area can be calculated directly by the definite integral of $y$ with respect to $x$ from $x=1$ to $x=5$.

Area (A) $= \int_{1}^{5} y \, dx$

Substitute $y = x^4$:

A $= \int_{1}^{5} x^4 \, dx$

Using the power rule of integration $\int x^n \, dx = \frac{x^{n+1}}{n+1}$:

A $= \left[ \frac{x^{4+1}}{4+1} \right]_{1}^{5}$

A $= \left[ \frac{x^5}{5} \right]_{1}^{5}$

Evaluate the expression at the limits of integration:

A $= \frac{5^5}{5} - \frac{1^5}{5}$

A $= \frac{3125}{5} - \frac{1}{5}$

A $= \frac{3125 - 1}{5}$

A $= \frac{3124}{5}$

Conclusion:

The area under the given curve and lines is $\frac{3124}{5}$ square units.

Given:

The curve $y = |x + 3|$, the interval $[-6, 0]$ on the x-axis.

To Sketch:

The graph of $y = |x + 3|$.

To Evaluate:

The definite integral $\int\limits_{−6}^0 |x + 3| \;dx$.

Solution:

The function $y = |x + 3|$ is defined as:

$y = \begin{cases} x + 3 & , & x + 3 \geq 0 \\ -(x + 3) & , & x + 3 < 0 \end{cases}$

which simplifies to:

$y = \begin{cases} x + 3 & , & x \geq -3 \\ -x - 3 & , & x < -3 \end{cases}$

Sketching the Graph:

The graph of $y = |x+3|$ is a V-shaped graph. The vertex of the V occurs where the expression inside the absolute value is zero, i.e., $x+3 = 0$, which gives $x = -3$. At $x=-3$, $y = |-3+3| = 0$. So, the vertex is at $(-3, 0)$.

For $x \geq -3$, the graph is the line $y = x+3$. This is a line passing through $(-3, 0)$ with a slope of 1. For example, when $x=0$, $y = 0+3 = 3$, so the point $(0, 3)$ is on the graph.

For $x < -3$, the graph is the line $y = -x-3$. This is a line passing through $(-3, 0)$ with a slope of -1. For example, when $x=-6$, $y = -(-6)-3 = 6-3 = 3$, so the point $(-6, 3)$ is on the graph.

The graph is symmetric about the vertical line $x = -3$.

Evaluating the Integral:

The integral $\int\limits_{−6}^0 |x + 3| \;dx$ represents the area between the curve $y = |x + 3|$ and the x-axis from $x = -6$ to $x = 0$.

Since the point where the definition of $|x+3|$ changes is $x=-3$, and this point lies within the interval of integration $[-6, 0]$, we split the integral into two parts:

$\int_{-6}^0 |x + 3| \, dx = \int_{-6}^{-3} |x + 3| \, dx + \int_{-3}^0 |x + 3| \, dx$

Using the definition of $|x+3|$:

For $x \in [-6, -3]$, $|x+3| = -(x+3) = -x - 3$.

For $x \in [-3, 0]$, $|x+3| = x+3$.

So the integral becomes:

$\int_{-6}^0 |x + 3| \, dx = \int_{-6}^{-3} (-x - 3) \, dx + \int_{-3}^0 (x + 3) \, dx$

First integral: $\int_{-6}^{-3} (-x - 3) \, dx = \left[ -\frac{x^2}{2} - 3x \right]_{-6}^{-3}$

$ = \left( -\frac{(-3)^2}{2} - 3(-3) \right) - \left( -\frac{(-6)^2}{2} - 3(-6) \right) $

$ = \left( -\frac{9}{2} + 9 \right) - \left( -\frac{36}{2} + 18 \right) $

$ = \left( \frac{-9 + 18}{2} \right) - (-18 + 18) $

$ = \frac{9}{2} - 0 = \frac{9}{2} $

Second integral: $\int_{-3}^{0} (x + 3) \, dx = \left[ \frac{x^2}{2} + 3x \right]_{-3}^{0}$

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

$ = (0 + 0) - \left( \frac{9}{2} - 9 \right) $

$ = 0 - \left( \frac{9 - 18}{2} \right) $

$ = - \left( -\frac{9}{2} \right) = \frac{9}{2} $

Summing the two parts:

$\int_{-6}^0 |x + 3| \, dx = \frac{9}{2} + \frac{9}{2} = \frac{18}{2} = 9$

Alternate Method (Geometric Interpretation):

The integral represents the area of the region bounded by $y = |x+3|$, the x-axis, $x=-6$, and $x=0$. This region consists of two triangles.

Triangle 1: Bounded by $y=-x-3$, the x-axis, $x=-6$ and $x=-3$. Vertices are $(-6, 0)$, $(-3, 0)$, and $(-6, |-6+3|) = (-6, 3)$. Base length is $|-3 - (-6)| = 3$. Height is 3. Area $= \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$.

Triangle 2: Bounded by $y=x+3$, the x-axis, $x=-3$ and $x=0$. Vertices are $(-3, 0)$, $(0, 0)$, and $(0, |0+3|) = (0, 3)$. Base length is $|0 - (-3)| = 3$. Height is 3. Area $= \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$.

Total Area $= \text{Area}_1 + \text{Area}_2 = \frac{9}{2} + \frac{9}{2} = 9$.

Conclusion:

The value of the integral $\int\limits_{−6}^0 |x + 3| \;dx$ is 9.

Given:

The curve $y = \sin x$, the x-axis ($y=0$), and the lines $x = 0$ and $x = 2\pi$.

To Find:

The area bounded by the curve $y = \sin x$ and the x-axis between $x = 0$ and $x = 2\pi$.

Solution:

The area bounded by the curve $y = f(x)$ and the x-axis from $x=a$ to $x=b$ is given by the integral $\int_{a}^{b} |f(x)| \, dx$.

In this case, $f(x) = \sin x$, $a=0$, and $b=2\pi$. The integral is $\int_{0}^{2\pi} |\sin x| \, dx$.

We need to consider the intervals where $\sin x$ is positive and where it is negative in the interval $[0, 2\pi]$.

We know that $\sin x \geq 0$ for $x \in [0, \pi]$ and $\sin x \leq 0$ for $x \in [\pi, 2\pi]$.

So, we split the integral at $x = \pi$:

Area (A) $= \int_{0}^{\pi} |\sin x| \, dx + \int_{\pi}^{2\pi} |\sin x| \, dx$

Since $\sin x \geq 0$ on $[0, \pi]$, $|\sin x| = \sin x$ in this interval.

Since $\sin x \leq 0$ on $[\pi, 2\pi]$, $|\sin x| = -\sin x$ in this interval.

A $= \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} (-\sin x) \, dx$

A $= \int_{0}^{\pi} \sin x \, dx - \int_{\pi}^{2\pi} \sin x \, dx$

Now, we evaluate the definite integrals. The indefinite integral of $\sin x$ is $-\cos x$.

First integral: $\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi}$

$= (-\cos \pi) - (-\cos 0)$

$= (-(-1)) - (-1)$

$= 1 + 1 = 2$

Second integral: $\int_{\pi}^{2\pi} \sin x \, dx = [-\cos x]_{\pi}^{2\pi}$

$= (-\cos 2\pi) - (-\cos \pi)$

$= (-1) - (-(-1))$

$= -1 - 1 = -2$

Substitute these values back into the expression for the total area:

A $= (2) - (-2)$

A $= 2 + 2 = 4$

Conclusion:

The area bounded by the curve $y = \sin x$ and the x-axis between $x = 0$ and $x = 2\pi$ is $4$ square units.

Given:

The curve $y = x^3$, the x-axis ($y=0$), and the ordinates $x = –2$ and $x = 1$.

To Find:

The area bounded by the given curve, the x-axis, and the lines.

Solution:

The area bounded by the curve $y = f(x)$, the x-axis, and the lines $x=a$ and $x=b$ is given by $\int_{a}^{b} |f(x)| \, dx$.

In this case, $f(x) = x^3$, $a=-2$, and $b=1$. The area is $\int_{-2}^{1} |x^3| \, dx$.

We need to determine where $x^3$ is positive and where it is negative in the interval $[-2, 1]$.

The function $f(x) = x^3$ is negative for $x < 0$ and positive for $x > 0$. It is zero at $x=0$. The point $x=0$ lies within the interval $[-2, 1]$.

We split the integral into two parts at $x=0$:

Area (A) $= \int_{-2}^{0} |x^3| \, dx + \int_{0}^{1} |x^3| \, dx$

For $x \in [-2, 0]$, $x^3 \leq 0$, so $|x^3| = -x^3$.

For $x \in [0, 1]$, $x^3 \geq 0$, so $|x^3| = x^3$.

So the integral becomes:

A $= \int_{-2}^{0} (-x^3) \, dx + \int_{0}^{1} x^3 \, dx$

A $= -\int_{-2}^{0} x^3 \, dx + \int_{0}^{1} x^3 \, dx$

Now, we evaluate each definite integral. The indefinite integral of $x^3$ is $\frac{x^4}{4}$.

First part: $-\left[ \frac{x^4}{4} \right]_{-2}^{0} = -\left( \frac{0^4}{4} - \frac{(-2)^4}{4} \right)$

$= -\left( 0 - \frac{16}{4} \right) = -\left( -4 \right) = 4$

Second part: $\left[ \frac{x^4}{4} \right]_{0}^{1} = \left( \frac{1^4}{4} - \frac{0^4}{4} \right)$

$= \left( \frac{1}{4} - 0 \right) = \frac{1}{4}$

The total area is the sum of these two values:

A $= 4 + \frac{1}{4} = \frac{16}{4} + \frac{1}{4} = \frac{17}{4}$

Conclusion:

The area bounded by the curve $y = x^3$, the x-axis and the ordinates $x = – 2$ and $x = 1$ is $\frac{17}{4}$ square units.

Comparing with the given options:

(A) -9

(B) $\frac{−15}{4}$

(C) $\frac{15}{4}$

(D) $\frac{17}{4}$

The correct answer is (D).

The curve $y = x | x |$, the x-axis ($y=0$), and the ordinates $x = – 1$ and $x = 1$.

To Find:

The area bounded by the given curve, the x-axis, and the lines.

Solution:

The given curve is $y = x | x |$. We can define this function piecewise:

$y = \begin{cases} x \cdot x = x^2 & , & \text{if } x \geq 0 \\ x \cdot (-x) = -x^2 & , & \text{if } x < 0 \end{cases}$

The area bounded by the curve $y = f(x)$ and the x-axis from $x=a$ to $x=b$ is given by the definite integral $\int\limits_{a}^{b} |f(x)| \, dx$.

Here, $f(x) = x|x|$, and the interval is $[-1, 1]$. The area (A) is $\int\limits_{-1}^{1} |x|x|| \, dx$.

We need to consider the definition of $|x|x||$ over the interval $[-1, 1]$. The point where the definition of $|x|$ changes is $x=0$, which is within the integration interval.

For $x \in [-1, 0)$: $x < 0$, so $|x| = -x$. The function is $y = x|x| = x(-x) = -x^2$. Since $-x^2 \leq 0$ in this interval, the area contribution is $\int\limits_{-1}^{0} |-x^2| \, dx$. We know that $|-x^2| = x^2$ for all real $x$. So this integral is $\int\limits_{-1}^{0} x^2 \, dx$.

For $x \in [0, 1]$: $x \geq 0$, so $|x| = x$. The function is $y = x|x| = x(x) = x^2$. Since $x^2 \geq 0$ in this interval, the area contribution is $\int\limits_{0}^{1} |x^2| \, dx = \int\limits_{0}^{1} x^2 \, dx$.

So, the total area is the sum of these two integrals:

A $= \int\limits_{-1}^{0} x^2 \, dx + \int\limits_{0}^{1} x^2 \, dx$

Now, we evaluate the definite integrals. The indefinite integral of $x^2$ is $\frac{x^3}{3}$.

$\int\limits_{-1}^{0} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{0} = \frac{0^3}{3} - \frac{(-1)^3}{3} = 0 - \frac{-1}{3} = \frac{1}{3}$

$\int\limits_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} - 0 = \frac{1}{3}$

Adding the areas:

A $= \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

Alternate Method (Using Property of Definite Integral):

We need to calculate the area $\int\limits_{-1}^{1} |x|x|| \, dx$. As shown in the main solution, $|x|x|| = x^2$ for all $x \in [-1, 1]$.

So, Area (A) $= \int\limits_{-1}^{1} x^2 \, dx$

The function $f(x) = x^2$ is an even function, since $f(-x) = (-x)^2 = x^2 = f(x)$.

For an even function $f(x)$, $\int\limits_{-a}^{a} f(x) \, dx = 2 \int\limits_{0}^{a} f(x) \, dx$.

Here, $a=1$ and $f(x) = x^2$.

A $= 2 \int\limits_{0}^{1} x^2 \, dx$

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

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

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

A $= 2 \times \frac{1}{3} = \frac{2}{3}$

Conclusion:

The area bounded by the curve y = x | x | , x-axis and the ordinates x = – 1 and x = 1 is $\frac{2}{3}$ square units.

Comparing with the given options:

(A) 0

(B) $\frac{1}{3}$

(C) $\frac{2}{3}$

(D) $\frac{4}{3}$

The correct answer is (C).