CBSE Class 12 Mathematics Application of Integrals MCQs Set 07

Answer: (d) \( \frac{5}{2} \)

Explanation:
1. y = x is below the axis on [−1, 0) and above it on (0, 2].
2. Area \( = \left|\int_{-1}^{0}x\,dx\right| + \int_{0}^{2}x\,dx = \frac{1}{2} + 2 \).
3. \( = \frac{5}{2} \) sq. units.

Teacher's Note:
1. Integrating directly from −1 to 2 gives 3/2, which is a common wrong choice; split at x = 0.
2. The region is two triangles, with areas ½ × 1 × 1 and ½ × 2 × 2, so no integration is really needed.

Answer: (b) \( \frac{64}{3} \)

Explanation:
1. The region lies between the y-axis and the curve x = y², from y = 0 to y = 4.
2. Area \( = \int_{0}^{4} y^2\,dy = \left[\frac{y^3}{3}\right]_{0}^{4} \).
3. \( = \frac{64}{3} \) sq. units.

Teacher's Note:
1. Use horizontal strips when the region is bounded by the y-axis and a horizontal line.
2. Check: the enclosing rectangle is 16 × 4 = 64, and the parabolic region next to the axis is one-third of it.

Answer: (d) 4π sq. units

Explanation:
1. By parts, \( \int x\sin x\,dx = -x\cos x + \sin x \).
2. On [0, π], x sin x ≥ 0: \( [-x\cos x + \sin x]_{0}^{\pi} = \pi \).
3. On [π, 2π], x sin x ≤ 0: \( [-x\cos x + \sin x]_{\pi}^{2\pi} = -2\pi - \pi = -3\pi \).
4. Area = π + 3π = 4π sq. units.

Teacher's Note:
1. Split the interval at x = π, where x sin x changes sign, and add the moduli.
2. The second arch is larger because the factor x is bigger there, so the parts do not cancel to 0 but give π and 3π.

Answer: (b) 4

Explanation:
1. cos x ≥ 0 on [0, π/2] and [3π/2, 2π], and ≤ 0 on [π/2, 3π/2].
2. \( \int_{0}^{\pi/2}\cos x\,dx = 1 \), \( \int_{\pi/2}^{3\pi/2}\cos x\,dx = -2 \), \( \int_{3\pi/2}^{2\pi}\cos x\,dx = 1 \).
3. Area = 1 + 2 + 1 = 4 sq. units.

Teacher's Note:
1. Over a full period the integral of cos x is 0, but the area is 4; split wherever the curve crosses the axis.
2. By symmetry, area = 4 × \( \int_{0}^{\pi/2}\cos x\,dx \) = 4.

Answer: (b) 2

Explanation:
1. sin x ≥ 0 on [0, π].
2. Area \( = \int_{0}^{\pi}\sin x\,dx = [-\cos x]_{0}^{\pi} = 1 + 1 \).
3. = 2 sq. units.

Teacher's Note:
1. The arch is symmetric about x = π/2, so area = \( 2\int_{0}^{\pi/2}\sin x\,dx = 2 \times 1 \).
2. Remember cos π = −1; using cos π = 1 gives the wrong answer 0.

Answer: (d) \( \frac{76\sqrt{a}}{3} \)

Explanation:
1. The region between the upper branch \( y = 2\sqrt{a}\sqrt{x} \), the axis y = 0 and the ordinates x = 4, x = 9 is taken.
2. Area \( = \int_{4}^{9} 2\sqrt{a}\,x^{1/2}\,dx = 2\sqrt{a}\cdot\frac{2}{3}\left[x^{3/2}\right]_{4}^{9} = \frac{4\sqrt{a}}{3}(27 - 8) \).
3. \( = \frac{76\sqrt{a}}{3} \) sq. units.

Teacher's Note:
1. \( 9^{3/2} = 27 \) and \( 4^{3/2} = 8 \); take the square root first and then cube.
2. \( \int \sqrt{x}\,dx = \frac{2}{3}x^{3/2} \); the factor \( 2\sqrt{a} \) comes out as a constant.

Answer: (b) \( \frac{28\sqrt{2}}{3} \) sq. units

Explanation:
1. The parabola y² = 2x is symmetric about the x-axis, and the region between x = 1 and x = 4 lies on both sides.
2. Area \( = 2\int_{1}^{4}\sqrt{2x}\,dx = 2\sqrt{2}\cdot\frac{2}{3}\left[x^{3/2}\right]_{1}^{4} = \frac{4\sqrt{2}}{3}(8 - 1) \).
3. \( = \frac{28\sqrt{2}}{3} \) sq. units.

Teacher's Note:
1. For a region bounded by the parabola itself (not its axis), take both halves; the upper half alone would give \( \frac{14\sqrt{2}}{3} \).
2. Use symmetry: compute the part above the x-axis and double it.

Answer: (c) \( \frac{8}{3}a^2 \) sq. units

Explanation:
1. The latus rectum is the line x = a; the region is symmetric about the x-axis.
2. Area \( = 2\int_{0}^{a} 2\sqrt{a}\sqrt{x}\,dx = 4\sqrt{a}\cdot\frac{2}{3}a^{3/2} \).
3. \( = \frac{8}{3}a^2 \) sq. units.

Teacher's Note:
1. The region is \( \frac{2}{3} \) of the rectangle of width a and height 4a (the length of the latus rectum): \( \frac{2}{3} \times 4a^2 = \frac{8}{3}a^2 \).
2. Forgetting the factor 2 for the lower half gives \( \frac{4}{3}a^2 \).

Answer: (a) \( \frac{4}{3}a^2 \)

Explanation:
1. The region bounded by the x-axis lies on one side of it, so take the upper branch \( y = 2\sqrt{ax} \).
2. Area \( = \int_{0}^{a} 2\sqrt{a}\,x^{1/2}\,dx = 2\sqrt{a}\cdot\frac{2}{3}a^{3/2} \).
3. \( = \frac{4}{3}a^2 \).

Teacher's Note:
1. Compare with the previous type: when the x-axis is a boundary, only one half of the parabola is used.
2. This is half the area cut off by the latus rectum, \( \frac{8}{3}a^2 \).

Answer: (a) πab sq. units

Explanation:
1. By symmetry, area = 4 × area in the first quadrant, where \( y = \frac{b}{a}\sqrt{a^2 - x^2} \).
2. \( \int_{0}^{a}\sqrt{a^2 - x^2}\,dx = \left[\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_{0}^{a} = \frac{\pi a^2}{4} \).
3. Area \( = 4\cdot\frac{b}{a}\cdot\frac{\pi a^2}{4} = \pi ab \) sq. units.

Teacher's Note:
1. Check: when b = a the ellipse becomes a circle and πab becomes πa².
2. \( \int_{0}^{a}\sqrt{a^2 - x^2}\,dx \) is a quarter circle of radius a, so its value \( \frac{\pi a^2}{4} \) can be written without integrating.

Answer: (b) \( 9\sec^{-1}(3) - \sqrt{8} \)

Explanation:
1. The smaller segment lies between x = 1 and x = 3; by symmetry, area \( = 2\int_{1}^{3}\sqrt{9 - x^2}\,dx \).
2. \( \int_{1}^{3}\sqrt{9 - x^2}\,dx = \left[\frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2}\sin^{-1}\frac{x}{3}\right]_{1}^{3} = \frac{9\pi}{4} - \sqrt{2} - \frac{9}{2}\sin^{-1}\frac{1}{3} \).
3. Area \( = 9\left(\frac{\pi}{2} - \sin^{-1}\frac{1}{3}\right) - 2\sqrt{2} = 9\cos^{-1}\frac{1}{3} - \sqrt{8} \).
4. Since \( \cos^{-1}\frac{1}{3} = \sec^{-1}3 \), area \( = 9\sec^{-1}3 - \sqrt{8} \).

Teacher's Note:
1. Use \( \frac{\pi}{2} - \sin^{-1}x = \cos^{-1}x \) and \( \cos^{-1}\frac{1}{x} = \sec^{-1}x \) to match the form of the options.
2. Check: 9 sec⁻¹3 − √8 ≈ 8.25, less than half the circle (≈ 14.1), as a smaller segment must be.

Answer: (a) \( \int_{0}^{2} \sqrt{2x - x^2}\,dx \)

Explanation:
1. The circle has centre (1, 0) and radius 1, so x runs from 0 to 2.
2. Upper half: \( y = \sqrt{1 - (x - 1)^2} = \sqrt{2x - x^2} \).
3. Area \( = \int_{0}^{2}\sqrt{2x - x^2}\,dx \) (its value is \( \frac{\pi}{2} \), so (d) \( \frac{\pi}{4} \) is wrong).

Teacher's Note:
1. The limits must cover the whole diameter, from 1 − 1 to 1 + 1; options (b) and (c) give only a quarter circle each.
2. Area of a semicircle of radius 1 is π/2, a quick check of any integral you set up.

Answer: (d) 1

Explanation:
1. y = |x − 2| is 2 − x on [1, 2] and x − 2 on [2, 3].
2. Area \( = \int_{1}^{2}(2 - x)\,dx + \int_{2}^{3}(x - 2)\,dx = \frac{1}{2} + \frac{1}{2} \).
3. = 1 sq. unit.

Teacher's Note:
1. The region is two triangles, each with base 1 and height 1, so the area is 2 × ½ = 1.
2. Split a modulus function at the point where the expression inside becomes zero.

Answer: (c) 16 log 2 + 3

Explanation:
1. y(x − 2) = 3x + 10, so \( y = \frac{3x + 10}{x - 2} = 3 + \frac{16}{x - 2} \), positive for 3 ≤ x ≤ 4.
2. Area \( = \int_{3}^{4}\left(3 + \frac{16}{x - 2}\right)dx = [3x + 16\log(x - 2)]_{3}^{4} \).
3. = 3 + 16(log 2 − log 1) = 16 log 2 + 3.

Teacher's Note:
1. Divide the numerator by the denominator first to split the rational function into a constant plus a simple fraction.
2. Check the curve is above the x-axis on the interval before integrating; here y ≥ 11.

Answer: (a) a = 3, b = −1

Explanation:
1. Through (1, 2): a + b = 2.
2. Area \( = \int_{0}^{4}(a\sqrt{x} + bx)\,dx = \frac{2a}{3}\cdot 8 + \frac{b}{2}\cdot 16 = \frac{16a}{3} + 8b = 8 \), i.e., 2a + 3b = 3.
3. Solving with a + b = 2: b = −1, a = 3.
4. Check: y = 3√x − x ≥ 0 on [0, 4] (it is 2 at x = 4), so the integral is indeed the area.

Teacher's Note:
1. Two unknowns need two equations: one from the point on the curve and one from the area condition.
2. Testing the options in a + b = 2 is a quick check: only (a) satisfies it.

Answer: (a) 3(x − 1) cos (3x + 4) + sin (3x + 4)

Explanation:
1. \( \int_{1}^{b} f(x)\,dx = (b - 1)\sin(3b + 4) \).
2. Differentiating both sides with respect to b: \( f(b) = \sin(3b + 4) + 3(b - 1)\cos(3b + 4) \).
3. So f(x) = 3(x − 1) cos(3x + 4) + sin(3x + 4).

Teacher's Note:
1. If the area up to x = b is given as a function of b, its derivative with respect to b gives the curve (Fundamental Theorem of Calculus).
2. f(x) must not contain the letter b; options (b) and (c) mix the variable and the limit.

Answer: (d) \( \frac{104a^2}{3} \)

Explanation:
1. Area \( = \int y\,dx \) with x = at², dx = 2at dt, and t from 1 to 3 (y > 0).
2. Area \( = \int_{1}^{3} 2at\cdot 2at\,dt = 4a^2\int_{1}^{3}t^2\,dt = 4a^2\cdot\frac{27 - 1}{3} \).
3. \( = \frac{104a^2}{3} \).

Teacher's Note:
1. For a curve in parametric form, change dx to x′(t) dt and the x-limits to t-limits.
2. A common mistake is to integrate y dt directly and forget the factor dx/dt = 2at.

Answer: (a) \( \frac{7}{120} \)

Explanation:
1. y′ = 8x³ − 2x = 2x(4x² − 1) = 0 gives x = 0, ±½; y″ = 24x² − 2 > 0 at x = ±½, so the minima are at x = ±½.
2. y = x²(2x² − 1) ≤ 0 for |x| ≤ ½, so the region lies below the x-axis.
3. Area \( = 2\int_{0}^{1/2}(x^2 - 2x^4)\,dx = 2\left(\frac{1}{24} - \frac{1}{80}\right) = 2\cdot\frac{7}{240} \).
4. \( = \frac{7}{120} \).

Teacher's Note:
1. Find the extreme points with the first derivative and confirm minima with the second derivative before setting the limits.
2. The curve is even, so integrate from 0 to ½ and double.

Answer: (a) \( \frac{5}{6} \)

Explanation:
1. f′(x) = 2x + 1 gives f(x) = x² + x + C; f(1) = 2 gives C = 0, so y = x² + x.
2. The curve meets the x-axis at x = 0 (and x = −1); the region with the line x = 1 is from x = 0 to x = 1.
3. Area \( = \int_{0}^{1}(x^2 + x)\,dx = \frac{1}{3} + \frac{1}{2} = \frac{5}{6} \).

Teacher's Note:
1. First find the curve from its slope and the given point; the constant of integration matters.
2. The lower limit is the point where the curve cuts the x-axis nearest to x = 1, here x = 0.

Answer: (b) \( 2\sqrt{2} \)

Explanation:
1. Total area \( = \int_{2}^{4}\left(1 + \frac{8}{x^2}\right)dx = \left[x - \frac{8}{x}\right]_{2}^{4} = (4 - 2) - (2 - 4) = 4 \).
2. Need \( \int_{2}^{a}\left(1 + \frac{8}{x^2}\right)dx = 2 \): \( a - \frac{8}{a} - (2 - 4) = 2 \), so \( a - \frac{8}{a} = 0 \).
3. a² = 8, and since 2 < a < 4, \( a = 2\sqrt{2} \).

Teacher's Note:
1. Set the area up to a equal to half the total area, rather than equating two integrals.
2. The dividing ordinate must lie between 2 and 4, which rules out 8, 2 and √2 at once.

Answer: (a) \( \frac{17}{27} \)

Explanation:
1. x² = 2x(1 − x) at x = 2/3, and (1 − x)² = 2x(1 − x) at x = 1/3; so f = (1 − x)² on [0, 1/3], 2x(1 − x) on [1/3, 2/3], x² on [2/3, 1].
2. \( \int_{0}^{1/3}(1 - x)^2dx = \frac{19}{81} \) and \( \int_{2/3}^{1}x^2dx = \frac{19}{81} \).
3. \( \int_{1/3}^{2/3}2x(1 - x)\,dx = \left[x^2 - \frac{2x^3}{3}\right]_{1/3}^{2/3} = \frac{20}{81} - \frac{7}{81} = \frac{13}{81} \).
4. Area \( = \frac{19 + 13 + 19}{81} = \frac{51}{81} = \frac{17}{27} \).

Teacher's Note:
1. For a maximum of several functions, find the crossing points and test one value in each piece to see which function is on top.
2. The graph is symmetric about x = ½, so the two outer pieces have equal areas.

Answer: (a) \( \frac{16}{3} \) sq. units

Explanation:
1. For −1 ≤ x ≤ 1, y = 2 − x² ≥ 1 > 0; for 1 < x ≤ 2, y = 2x − 1 > 0.
2. \( \int_{-1}^{1}(2 - x^2)\,dx = 4 - \frac{2}{3} = \frac{10}{3} \).
3. \( \int_{1}^{2}(2x - 1)\,dx = [x^2 - x]_{1}^{2} = 2 \).
4. Area \( = \frac{10}{3} + 2 = \frac{16}{3} \) sq. units.

Teacher's Note:
1. Split a piecewise function at the point where its rule changes and integrate each rule on its own interval.
2. Option (b), 10/3, is only the first piece; do not forget the straight-line part.

Answer: (c) \( \frac{2}{3} \)

Explanation:
1. Take y² = 4ax and the double ordinate x = h, with ends \( (h, \pm 2\sqrt{ah}) \).
2. Area cut off \( = 2\int_{0}^{h}2\sqrt{a}\sqrt{x}\,dx = \frac{8}{3}\sqrt{a}\,h^{3/2} \).
3. Rectangle = (length of double ordinate) × (distance from vertex) \( = 4\sqrt{ah}\cdot h = 4\sqrt{a}\,h^{3/2} \).
4. Ratio \( = \frac{8/3}{4} = \frac{2}{3} \).

Teacher's Note:
1. This is Archimedes' result: a parabolic segment is two-thirds of its bounding rectangle, a useful check for many parabola areas.
2. The ratio does not depend on a or h, so you can verify it with a = h = 1.