CBSE Class 12 Mathematics Application of Integrals MCQs Set 04

Question. Area of the region bounded by the curve \(y^2 = 4x\) and the X-axis between \(x = 0\) and \(x = 1\) is
(a) \(\frac{2}{3}\)
(b) \(\frac{8}{3}\)
(c) 3
(d) \(\frac{4}{3}\)
Answer: (b) \(\frac{8}{3}\)

Question. The area bounded by the curve \(y = \sqrt{x}\), Y-axis and between the lines \(y = 0\) and \(y = 3\) is
(a) \(2\sqrt{3}\)
(b) 27
(c) 9
(d) 3
Answer: (c) 9

Question. The area under the curve \(y = x^2\) between the line \(x = 0\) and \(x = k\) is 9 sq units. Which of the following could be the value of \(k\)?
(a) 3
(b) 4.5
(c) 9
(d) 27
Answer: (a) 3

Question. Shown below is the graph of \(-x = y^2 - 3\).
Ravi says that the area under the curve in the first quadrant can be found as \(\int_0^{\sqrt{3}} \sqrt{3 - x} dx\).
Kanika says that the area under the curve in the first quadrant can be found as \(\int_0^{\sqrt{3}} (3 - y^2) dy\).
Who is correct?
(a) Only Ravi
(b) Only Kanika
(c) Both Ravi and Kanika
(d) Neither Ravi nor Kanika
Answer: (c) Both Ravi and Kanika

Question. The area of region bounded by the \(y = x|x|\) and \(x = -2\), \(x = 2\) and X-axis is
(a) \(\frac{1}{3}\) sq units
(b) \(\frac{16}{3}\) sq units
(c) \(\frac{4}{3}\) sq units
(d) \(\frac{7}{3}\) sq units
Answer: (b) \(\frac{16}{3}\) sq units

Question. The area bounded by the curve \(y = \cos x\) and X-axis and the ordinates \(x = 0\) and \(x = \pi\) is
(a) 4 sq units
(b) 3 sq units
(c) 2 sq units
(d) 1 sq unit
Answer: (c) 2 sq units

Question. Shown below are partial graphs of two distinct functions, \(f(y)\) and \(g(y)\). The region between them is shaded.
Which expression gives the area of this shaded region?
(a) \(\int_a^b |f(y) - g(y)| dy\)
(b) \(\int_a^b |f(x) - g(x)| dx\)
(c) \(\int_c^d |f(y) - g(y)| dy\)
(d) \(\int_c^d |f(x) + g(x)| dx\)
Answer: (c) \(\int_c^d |f(y) - g(y)| dy\)

Question. The area of the shaded region represented by the curves \(y = x^2\), \(0 \le x \le 2\) and Y-axis is given by
(a) \(\int_0^2 x^2 dx\)
(b) \(\int_0^2 \sqrt{y} dy\)
(c) \(\int_0^4 x^2 dx\)
(d) \(\int_0^4 \sqrt{y} dy\)
Answer: (d) \(\int_0^4 \sqrt{y} dy\)

Question. Area enclosed by the curve \(x^2 + y^2 = 9\) is
(a) \(9\pi\) sq units
(b) \(3\pi\) sq units
(c) 9 sq units
(d) None of the options
Answer: (a) \(9\pi\) sq units

Question. The area of the region bounded by the curve \(y = \sqrt{16 - x^2}\) and X-axis is
(a) \(8\pi\) sq units
(b) \(64\pi\) sq units
(c) \(4\pi\) sq units
(d) None of the options
Answer: (a) \(8\pi\) sq units

Question. The area (in sq units) of the region bounded by the curve \(y = x\), X-axis, \(x = 0\) and \(x = 2\) is
(a) \(\frac{3}{2}\)
(b) \(\frac{1}{2}\log 2\)
(c) 2
(d) 4
Answer: (c) 2

Question. If a curve \(y = a\sqrt{x} + bx\) passes through the point \((1, 2)\) and the area bounded by the curve, line \(x = 4\) and X-axis is 8 sq units, then
(a) \(a = 3, b = -1\)
(b) \(a = 3, b = 1\)
(c) \(a = -3, b = 1\)
(d) \(a = -3, b = -1\)
Answer: (a) \(a = 3, b = -1\)

Question. The area bounded by the curve \(y = x|x|\), X-axis and the coordinates \(x = -1\) and \(x = 1\) is given by
(a) 0
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{4}{3}\)
Answer: (c) \(\frac{2}{3}\)

Question. 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
(a) \(\pi\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{4}\)
Answer: (a) \(\pi\)

Question. Area of the region bounded by the curve \(y^2 = 4x\), Y-axis and the line \(y = 3\) is
(a) 2
(b) \(\frac{9}{4}\)
(c) \(\frac{9}{3}\)
(d) \(\frac{9}{2}\)
Answer: (b) \(\frac{9}{4}\)

Assertion Reason Based Questions

Question. Assertion (A) The area bounded by \(y^2 = 4x\) and \(y = x\) is \(\frac{8}{3}\) sq units.
Reason (R) The area bounded by \(y^2 = 4ax\) and \(y = mx\) is \(\frac{8a^2}{3m^3}\) sq units.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) The area bounded by the curve \(y = \sin x\) between \(x = 0\) and \(x = 4\pi\) is 4 sq units.
Reason (R) \(\int_0^{\pi/2} \sin x dx = 1\).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A) Area enclosed by the circle \(x^2 + y^2 = 64\) is equal to \(64\pi\) sq units.
Reason (R) Area enclosed by the circle \(x^2 + y^2 = r^2\) is \(\pi r^2\).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) Area bounded between \(x^2 + y^2 = 4\) and \(x + y = 2\) is \((\pi - 2)\) sq units.
Reason (R) The area bounded by circle \(x^2 + y^2 = a^2\) and line \(x + y = a\) is represented by \(\int_0^a \sqrt{a^2 - x^2} dx - \int_0^a (a - x) dx\).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (a) Both A and R are correct; R is the correct explanation of A.

Question. Assertion (A) Area of the region given by \(\{(x, y) : y^2 \le 6x, 2 \le x \le 5, x, y \ge 0\}\) is \(\frac{21}{2}\) sq unit.
Reason (R) Area under a curve \(x = f(y)\) lying to the right of Y-axis between \(y = a\) and \(y = b\) is given by \(\int_a^b f(y) dy\).
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (d) R is correct; A is incorrect.

Question. Assertion (A) Area bounded by the curve \(y = |x - 1|\) and \(y = 1\) is 1 sq unit.
Reason (R) The area bounded by the curve \(y = |x|\), the X-axis and between \(x = -2\) and \(x = 1\) is \(\frac{5}{2}\) sq units.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Question. Assertion (A) Area of the lines \(y = \sqrt{3}x\) and \(y = 0\) and \(x = 4\) in first quadrant is \(8\sqrt{3}\) sq units.
Reason (R) Area of the lines \(y = \sqrt{3}x\) and \(x = 0\) and \(y = 4\) in first quadrant is \(\frac{8}{\sqrt{3}}\) sq units.
(a) Both A and R are correct; R is the correct explanation of A.
(b) Both A and R are correct; R is not the correct explanation of A.
(c) A is correct; R is incorrect.
(d) R is correct; A is incorrect.
Answer: (b) Both A and R are correct; R is not the correct explanation of A.

Case Study Based Questions - I

A child cut a pizza with a knife. Pizza is circular in shape which is represented by \(x^2 + y^2 = 4\) and sharp edge of knife represents a straight line given by \(x = \sqrt{3}y\). Based on the above information, answer the following questions.

Question. Find the point of intersection of the edge of knife and pizza.
Answer: For the point of intersection, solve the equations: \[x^2 + y^2 = 4 \quad \text{--- (i)}\] \[x = \sqrt{3}y \quad \text{--- (ii)}\] Substituting (ii) in (i): \[(\sqrt{3}y)^2 + y^2 = 4 \implies 3y^2 + y^2 = 4 \implies 4y^2 = 4 \implies y^2 = 1 \implies y = \pm 1\] For \(y = 1\), \(x = \sqrt{3}\).
For \(y = -1\), \(x = -\sqrt{3}\).
Thus, the points of intersection are \(A(\sqrt{3}, 1)\) and \(B(-\sqrt{3}, -1)\).

Question. Draw the graph of both circle and line and hence shade the smaller area bounded by edge of knife and pizza in the first quadrant.
Answer: Plotting the circle \(x^2 + y^2 = 4\) (radius 2, center at the origin) and the line \(x = \sqrt{3}y\) passing through the origin. The smaller area enclosed between the line, the circle, and the X-axis in the first quadrant is shaded. The intersection point in the first quadrant is \(A(\sqrt{3}, 1)\).

Question. Using integration find the area of that smaller part.
Answer: The area of the shaded part is computed as: \[\text{Area} = \int_0^{\sqrt{3}} y_{\text{line}} dx + \int_{\sqrt{3}}^2 y_{\text{circle}} dx\] \[= \int_0^{\sqrt{3}} \frac{x}{\sqrt{3}} dx + \int_{\sqrt{3}}^2 \sqrt{4 - x^2} dx\] \[= \frac{1}{\sqrt{3}} \left[ \frac{x^2}{2} \right]_0^{\sqrt{3}} + \left[ \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right) \right]_{\sqrt{3}}^2\] \[= \frac{1}{\sqrt{3}} \left(\frac{3}{2}\right) + \left[ \left(0 + 2\sin^{-1}(1)\right) - \left(\frac{\sqrt{3}}{2}\sqrt{1} + 2\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\right) \right]\] \[= \frac{\sqrt{3}}{2} + \left[ 2\left(\frac{\pi}{2}\right) - \frac{\sqrt{3}}{2} - 2\left(\frac{\pi}{3}\right) \right]\] \[= \frac{\sqrt{3}}{2} + \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3} = \frac{\pi}{3} \text{ sq units}\]

Question. If child cut the pizza into four equal part. Find the area of each part.
Answer: If the pizza is cut into four equal parts, each part represents a quadrant of the circle. The area of each part is: \[\text{Area} = \int_0^2 \sqrt{4 - x^2} dx\] \[= \left[ \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_0^2\] \[= 2 \sin^{-1}(1) = 2 \times \frac{\pi}{2} = \pi \text{ sq units}\]

Question. Find the area of whole pizza.
Answer: The area of the whole pizza is: \[\text{Area} = 4 \times (\text{area of one quadrant}) = 4\pi \text{ sq units}\]

Case Study Based Questions - II

Location of three houses of a society is represented by the points \(A(-1, 0)\), \(B(1, 3)\) and \(C(3, 2)\) as shown in figure.

Question. Find the equation of line \(AB\).
Answer: The equation of line passing through \(A(-1, 0)\) and \(B(1, 3)\) is: \[y - 0 = \frac{3 - 0}{1 - (-1)}(x - (-1))\] \[y = \frac{3}{2}(x + 1)\]

Question. Find the equation of \(BC\).
Answer: The equation of line passing through \(B(1, 3)\) and \(C(3, 2)\) is: \[y - 3 = \frac{2 - 3}{3 - 1}(x - 1)\] \[y - 3 = -\frac{1}{2}(x - 1) \implies y = -\frac{x}{2} + \frac{7}{2}\]

Question. Find the area of \(\Delta ABE\).
Answer: The area is given by: \[\text{Area} = \int_{-1}^1 y_{AB} dx = \int_{-1}^1 \frac{3}{2}(x+1) dx\] \[= \frac{3}{2} \left[ \frac{x^2}{2} + x \right]_{-1}^1 = \frac{3}{2} \left[ \left(\frac{1}{2} + 1\right) - \left(\frac{1}{2} - 1\right) \right]\] \[= \frac{3}{2} [2] = 3 \text{ sq units}\]

Question. Find the area of trapezium \(EBCD\).
Answer: The area is given by: \[\text{Area} = \int_1^3 y_{BC} dx = \int_1^3 \left( -\frac{x}{2} + \frac{7}{2} \right) dx\] \[= \left[ -\frac{x^2}{4} + \frac{7x}{2} \right]_1^3 = \left( -\frac{9}{4} + \frac{21}{2} \right) - \left( -\frac{1}{4} + \frac{7}{2} \right)\] \[= \frac{33}{4} - \frac{13}{4} = \frac{20}{4} = 5 \text{ sq units}\]

Question. Find the area of \(\Delta ACD\).
Answer: Equation of line \(AC\) passing through \(A(-1, 0)\) and \(C(3, 2)\) is: \[y - 0 = \frac{2 - 0}{3 - (-1)}(x - (-1)) \implies y = \frac{1}{2}(x + 1)\] The area under line \(AC\) is: \[\text{Area} = \int_{-1}^3 y_{AC} dx = \int_{-1}^3 \frac{1}{2}(x + 1) dx\] \[= \frac{1}{2} \left[ \frac{x^2}{2} + x \right]_{-1}^3 = \frac{1}{2} \left[ \left(\frac{9}{2} + 3\right) - \left(\frac{1}{2} - 1\right) \right]\] \[= \frac{1}{2} [8] = 4 \text{ sq units}\]

Question. Find the area of \(\Delta ABC\).
Answer: The area is given by: \[\text{Area of } \Delta ABC = \text{Area of } \Delta ABE + \text{Area of trapezium } EBCD - \text{Area of } \Delta ACD\] \[= 3 + 5 - 4 = 4 \text{ sq units}\]

Case Study Based Questions - III

A Lorenz curve is used to graphically represent income inequality in a society. It was developed by Max Lorenz in 1905. In this curve, the percentile of the population according to their income is plotted on the X-axis and on the Y-axis, the percentage of cumulative income from that percentile of the population is plotted. e.g. In the graph below, the point (20, 4) denotes that people with more income than that of 20% of the total population contribute 4% of the total income for the country. Similarly, the point (64, 40) denotes that people with more income than that of 64% of the country's population contribute 40% of the total income for the country. On the graph, there is also a line of equality, given by the function \(g(x) = x\). The further away the Lorenz curve of a society is from the line of equality, the more unequal its income distribution is. In order to compare this data for multiple countries, the area under the Lorenz Curve is used to find the Gini coefficient (G), whose value ranges between 0 and 1. The closer G's value is to 1, the more unequal the income distribution of the society is. The Lorenz curve shown above can be approximated by the following function: \[f(x) = \frac{\sqrt{x}}{100} + \frac{(x-1)^2}{100}\]

Question. Find the area under the Lorenz curve from 0 to 100. Show your work.
Answer: The area under the Lorenz curve from 0 to 100 is: \[\text{Area} = \int_0^{100} f(x) dx = \int_0^{100} \left( \frac{\sqrt{x}}{100} + \frac{(x-1)^2}{100} \right) dx\] \[= \frac{1}{100} \int_0^{100} x^{1/2} dx + \frac{1}{100} \int_0^{100} (x^2 - 2x + 1) dx\] \[= \frac{1}{100} \left[ \frac{2}{3} x^{3/2} \right]_0^{100} + \frac{1}{100} \left[ \frac{x^3}{3} - x^2 + x \right]_0^{100}\] \[= \frac{1}{100} \left( \frac{2000}{3} \right) + \frac{1}{100} \left( \frac{1000000}{3} - 10000 + 100 \right)\] \[= \frac{20}{3} + \frac{10000}{3} - 100 + 1\] \[= \frac{10020}{3} - 99 = 3340 - 99 = 3241 \text{ sq units}\]

Question. The Gini coefficient is given by \(G = \frac{A}{A+B}\), where \(A\) is the area between the Lorenz curve and the line of equality and \(B\) is the area under the Lorenz curve. Find the Gini coefficient for the given Lorenz curve, using integration. Show your work.
Answer: The line of equality is represented by \(g(x) = x\).
The total area under the line of equality is: \[A + B = \int_0^{100} x dx = \left[ \frac{x^2}{2} \right]_0^{100} = \frac{10000}{2} = 5000 \text{ sq units}\] From the previous question, we have \(B = 3241\) sq units.
Therefore, the area \(A\) is: \[A = (A + B) - B = 5000 - 3241 = 1759 \text{ sq units}\] The Gini coefficient \(G\) is: \[G = \frac{A}{A+B} = \frac{1759}{5000} = 0.3518\]