CBSE Class 12 Mathematics HOTs Application of Integrals Set 02

Short Answer Type Questions

Question. Find the area of the region bounded by \( y^2 = 9x \), \( x = 2 \), \( x = 4 \) and the X-axis in the first quadrant.
Answer: Given equation of parabola is \( y^2 = 9x \).
Also, given ordinates \( x = 2 \) and \( x = 4 \).
The bounded region in \(\text{I}^{\text{st}}\) quadrant is \( BCFEB \).
Required area \( = \int_{2}^{4} y \, dx = 3 \int_{2}^{4} \sqrt{x} \, dx \)
Ans. \( 4(4 - \sqrt{2}) \) sq units.

Question. Using integration, evaluate the area of the region bounded by the curve \( y = x^2 \), the lines \( y = 1 \) and \( y = 3 \) and the Y-axis.
Answer: Given, curve \( y = x^2 \).
Required area \( = 2 \, (\text{Area of shaded region } ABCDA) \)
\( = 2 \int_{1}^{3} x \, dy = 2 \int_{1}^{3} \sqrt{y} \, dy \)
\( = \frac{4}{3} \left[ y^{3/2} \right]_{1}^{3} \)
\( = \frac{4}{3} \left[ (3)^{3/2} - 1^{3/2} \right] \)
\( = \frac{4}{3} [3\sqrt{3} - 1] \)
\( = \left[ 4\sqrt{3} - \frac{4}{3} \right] \) sq units.

Question. Using integration, find the area of the region bounded by \( y = mx \, (m > 0) \), \( x = 1 \), \( x = 2 \) and the X-axis.
Answer: Given, equation of curve \( y = mx \, (m > 0) \), \( x = 1, 2 \).
Required area \( = \text{Area of the region } ABCDA \)
\( = \int_{1}^{2} (mx) \, dx = m \left[ \frac{x^2}{2} \right]_{1}^{2} = m \left[ \frac{4}{2} - \frac{1}{2} \right] \)
\( = m \left( 2 - \frac{1}{2} \right) = \frac{3}{2}m \) sq units.

Question. Find the area of the smaller part of the circle \( x^2 + y^2 = a^2 \) cut-off by the line \( x = \frac{a}{\sqrt{2}} \).
Answer: Given equations of circle and line are
\( x^2 + y^2 = a^2 \) ...(i)
and \( x = \frac{a}{\sqrt{2}} \) ...(ii)
Clearly, required region is \( APBCA \), which is symmetrical about X-axis and the x-coordinate of point of intersection of curve and line is \( \frac{a}{\sqrt{2}} \).
Now, required area \( = \text{Area of region } APBCA \)
\( = 2 \, (\text{Area of region } APCA) \)
\( = 2 \int_{a/\sqrt{2}}^{a} y \, dx \)
\( = 2 \int_{a/\sqrt{2}}^{a} \sqrt{a^2 - x^2} \, dx \)
Ans. \( \frac{a^2(\pi - 2)}{4} \) sq units.

Question. Find the area of the region bounded by the curve \( y = x^2 \) and the line \( y = 4 \).
Answer: Required area \( = 2 \, (\text{Area of region } OABO) \)
\( = 2 \int_{0}^{4} x \, dy \)
Ans. \( \frac{32}{3} \) sq units.

Question. Find the area of the parabola \( y^2 = 4ax \) bounded by its latusrectum.
Answer: For parabola, \( y^2 = 4ax \).
Latusrectum is line \( x = a \).
Required area \( = \text{Area of } OBCAO = 2 \times \text{Area of } OCAO \)
\( = 2 \times \int_{0}^{a} y \, dx = 2 \int_{0}^{a} \sqrt{4ax} \, dx \)
Ans. \( \frac{8}{3}a^2 \) sq units.

Question. Find the area of the region bounded by \( y = -1 \), \( y = 2 \), \( x = y^3 \) and \( x = 0 \).
Answer: Required area \( = \left| \int_{-1}^{0} y^3 \, dy \right| + \left| \int_{0}^{2} y^3 \, dy \right| \)
Ans. \( \frac{17}{4} \) sq units.

Question. Find the area bounded by the curve \( y = x|x| \), X-axis and the ordinates \( x = -3 \) and \( x = 3 \).
Answer: Given equation of curve is
\( y = x|x| = \begin{cases} x^2, & \text{if } x \geq 0 \\ -x^2, & \text{if } x < 0 \end{cases} \)
The graph of above curve between the ordinates \( x = -3 \) and \( x = 3 \) is given below.
Clearly, shaded portion is the required region.
Required area \( = \text{Area of region } OALO + \text{Area of region } OBMO \)
\( = \int_{0}^{3} x^2 \, dx + \left| \int_{-3}^{0} -x^2 \, dx \right| \)
Ans. \( 18 \) sq units.

Question. Find the area of the region bounded by the line \( y = 5x + 2 \), the X-axis and the ordinates \( x = -2 \) and \( x = 2 \).
Answer: Given lines are
\( y = 5x + 2 \) ...(i)
\( y = 0 \) (X-axis) ...(ii)
\( x = -2 \) ...(iii)
and \( x = 2 \) ...(iv)
Table for \( y = 5x + 2 \) is:
\( \begin{array}{|c|c|c|} \hline x & 0 & -2/5 \\ \hline y & 2 & 0 \\ \hline \end{array} \)
The region bounded by \( y = 5x + 2 \), X-axis and the ordinates \( x = -2 \) and \( x = 2 \) is represented by the shaded portion.
Required area \( = \text{Area of region } EFDE + \text{Area of region } ABDA \)
\( = \left| \int_{-2}^{-2/5} y \, dx \right| + \int_{-2/5}^{2} y \, dx \)
Ans. \( \frac{104}{5} \) sq units.

Long Answer Type Questions

Question. Draw the graph of the curve \( y = |\sin x| \) and find the area bounded by the curve, X-axis and ordinates \( x = -\pi \) to \( 2\pi \).
Answer: We have, \( y = |\sin x| = \begin{cases} \sin x, & x \in (0, \pi), (2\pi, 3\pi), \dots \\ -\sin x, & x \in (-\pi, 0), (\pi, 2\pi), \dots \end{cases} \)
Now, required area \( = \text{Area of shaded region} \)
\( = 3 \times (\text{Area of one shaded region}) \)
[Since all parts are of equal areas]
\( = 3 \int_{0}^{\pi} y \, dx = 3 \int_{0}^{\pi} (\sin x) \, dx \)
Ans. \( 6 \) sq units.

Question. Draw a rough sketch of the curve \( y = |x-2| \). Find the area under the curve and line \( x = 0 \) and \( x = 4 \).
Answer: Required area \( = \int_{0}^{2} (2-x) \, dx + \int_{2}^{4} (x-2) \, dx \)
Ans. \( 4 \) sq units.

Question. Find the area bounded \( x^2 = 4y \) and \( y = 4 \) in the first quadrant.
Answer: Required area \( = \text{Area of } OABO = \int_{0}^{4} x \, dy \)
\( = \int_{0}^{4} 2\sqrt{y} \, dy = 2 \left[ \frac{y^{3/2}}{\frac{3}{2}} \right]_{0}^{4} \)
\( = \frac{4}{3} \left[ (4)^{3/2} - 0 \right] \)
\( = \frac{4}{3} \left[ (2^2)^{3/2} \right] = \frac{4}{3} [2^3] = \frac{32}{3} \) sq units.

Question. Find the area bounded by ellipse \( 9x^2 + 4y^2 = 36 \).
Answer: We have, \( 9x^2 + 4y^2 = 36 \)
\( \implies \frac{9x^2}{36} + \frac{4y^2}{36} = \frac{36}{36} \) [dividing by 36 both sides]
\( \implies \frac{x^2}{4} + \frac{y^2}{9} = 1 \)
The given curve is an ellipse with centre at \( (0, 0) \) and symmetric about both axes.
Since \( a^2 < b^2 \), it is a vertical ellipse.
Required area \( = 4 \times \text{ar (OABO)} \)
\( = 4 \int_{0}^{2} y \, dx \)
\( = 4 \int_{0}^{2} \frac{3}{2}\sqrt{4-x^2} \, dx \)
\( = 6 \int_{0}^{2} \sqrt{(2)^2 - x^2} \, dx \)
\( = 6 \left[ \frac{x}{2}\sqrt{2^2-x^2} + \frac{2^2}{2}\sin^{-1}\frac{x}{2} \right]_{0}^{2} \)
\( \left[ \dots \int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \right] \)
\( = 6 \left[ \frac{2}{2}\sqrt{2^2-2^2} + 2\sin^{-1}\frac{2}{2} - \left(0 + 2\sin^{-1}0\right) \right] \)
\( = 6 [0+2\sin^{-1}(1)-0] \)
\( = 6 \left( 2\sin^{-1}\left(\sin\frac{\pi}{2}\right) \right) = 6 \times 2 \times \frac{\pi}{2} \)
\( = 6\pi \) sq units.

Question. Sketch the graph of \( y = |x+3| \) and evaluate the area under the curve \( y = |x+3| \) above X-axis and between \( x = -6 \) to \( x = 0 \).
Answer: First, we sketch the graph of \( y = |x+3| \).
\( y = |x+3| = \begin{cases} x+3, & \text{if } x+3 \ge 0 \\ -(x+3), & \text{if } x+3 < 0 \end{cases} \)
\( \implies y = |x+3| = \begin{cases} x+3, & \text{if } x \ge -3 \\ -x-3, & \text{if } x < -3 \end{cases} \)
So, we have \( y = x+3 \) for \( x \ge -3 \) and \( y = -x-3 \) for \( x < -3 \).
A sketch of \( y = |x+3| \) is shown below.
Here, \( y = x+3 \) is the straight line which cuts X and Y-axes at \( (-3,0) \) and \( (0,3) \), respectively.
Thus, \( y = x+3 \) for \( x \ge -3 \) represents the part of line which lies on the right side of \( x = -3 \).
Similarly, \( y = -x-3, x < -3 \) represents the part of line \( y = -x-3 \), which lies on left side of \( x = -3 \).
Clearly, required area \( = \text{Area of region } ABPA + \text{Area of region } PCOP \)
\( = \int_{-6}^{-3} (-x-3) \, dx + \int_{-3}^{0} (x+3) \, dx \)
\( = \left[ -\frac{x^2}{2} - 3x \right]_{-6}^{-3} + \left[ \frac{x^2}{2} + 3x \right]_{-3}^{0} \)
\( = \left[ \left( -\frac{9}{2} + 9 \right) - \left( -18 + 18 \right) \right] + \left[ 0 - \left( \frac{9}{2} - 9 \right) \right] \)
\( = \frac{9}{2} + \frac{9}{2} = 9 \) sq units.

Question. Find the area bounded by the curve \( y = \cos x \) between \( x = 0 \) and \( x = 2\pi \).
Answer: Required area \( = \text{Area of } OABO + \text{Area of } BCDB + \text{Area of } DEFD \)
\( = \int_{0}^{\pi/2} y \, dx + \left| \int_{\pi/2}^{3\pi/2} y \, dx \right| + \int_{3\pi/2}^{2\pi} y \, dx \)
Ans. \( 4 \) sq units.

Question. Find the area bounded by the curve \( y = x^3 \), the X-axis and the ordinates \( x = -2 \) and \( x = 1 \).
Answer: Required area \( = \text{Area of } ABOA + \text{Area of } DCOD \)
\( = \left| \int_{-2}^{0} y \, dx \right| + \int_{0}^{1} y \, dx \)
Ans. \( \frac{17}{4} \) sq. units.

Question. The area between \( x = y^2 \) and \( x = 4 \) is divided into two equal parts by the line \( x = a \). Find the value of \( a \).
Answer: So, area \( (OABCDO) = 2 \times \text{Area (OADO)} \)
\( = 2 \int_{0}^{4} \sqrt{x} \, dx \)
\( = 2 \left( 2 \int_{0}^{a} \sqrt{x} \, dx \right) \)
\( \implies \frac{32}{3} = \frac{8}{3}(a)^{3/2} \)
\( \implies a = (4)^{2/3} \)

Question. Find the area bounded by the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) and the coordinates \( x = 0 \) and \( x = ae \), where \( b^2 = a^2(1-e^2) \) and \( e < 1 \).
Answer: Therefore, shaded area \( = 2 \times \text{Area (OABCO)} \)
\( = 2 \int_{0}^{ae} y \, dx \)
\( = 2 \int_{0}^{ae} \frac{b}{a}\sqrt{a^2-x^2} \, dx \)
\( = ab [e\sqrt{1-e^2} + \sin^{-1}e] \) sq units.