CBSE Class 12 Mathematics Application of Integrals MCQs Set 06

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

Explanation:
1. y = x² − 4x = x(x − 4) is negative for 0 < x < 2, so the region lies below the x-axis.
2. Area \( = \int_{0}^{2} (4x - x^2)\,dx = \left[2x^2 - \frac{x^3}{3}\right]_{0}^{2} \).
3. \( = 8 - \frac{8}{3} = \frac{16}{3} \) sq. units.

Teacher's Note:
1. When the curve is below the x-axis, integrate −y (or take the modulus of the integral); an area is never negative.
2. So option (b), the signed value of the integral, cannot be an area.

Answer: (c) \( \frac{32}{3} \) sq. units

Explanation:
1. 4x − x² = 0 gives x = 0 and x = 4, and the curve is above the x-axis between them.
2. Area \( = \int_{0}^{4} (4x - x^2)\,dx = \left[2x^2 - \frac{x^3}{3}\right]_{0}^{4} \).
3. \( = 32 - \frac{64}{3} = \frac{32}{3} \) sq. units.

Teacher's Note:
1. For a parabola y = −(x − α)(x − β), the area with the x-axis is \( \frac{(\beta - \alpha)^3}{6} \); here \( \frac{4^3}{6} = \frac{32}{3} \).
2. Always find where the curve meets the axis first; those points are the limits.

Answer: (a) \( \frac{125}{6} \)

Explanation:
1. 4 + 3x − x² = −(x − 4)(x + 1) = 0 gives x = −1 and x = 4; the curve is above the axis between them.
2. Area \( = \int_{-1}^{4} (4 + 3x - x^2)\,dx = \left[4x + \frac{3x^2}{2} - \frac{x^3}{3}\right]_{-1}^{4} \).
3. \( = \left(16 + 24 - \frac{64}{3}\right) - \left(-4 + \frac{3}{2} + \frac{1}{3}\right) = \frac{56}{3} + \frac{13}{6} = \frac{125}{6} \).

Teacher's Note:
1. Shortcut: for y = −(x − α)(x − β) the area is \( \frac{(\beta - \alpha)^3}{6} = \frac{5^3}{6} = \frac{125}{6} \).
2. Be careful with signs when substituting the negative lower limit x = −1.

Answer: (d) None of these

Explanation:
1. Area \( = \int_{0}^{4} \sqrt{3x + 4}\,dx = \left[\frac{2}{9}(3x + 4)^{3/2}\right]_{0}^{4} \).
2. At x = 4: (16)^{3/2} = 64; at x = 0: (4)^{3/2} = 8.
3. Area \( = \frac{2}{9}(64 - 8) = \frac{112}{9} \) sq. units, which is not in (a), (b), (c). Answer: None of these.

Teacher's Note:
1. For \( \int (ax + b)^n dx \) divide by a as well as by n + 1; forgetting the factor 1/3 here gives a wrong value.
2. Quick estimate: √(3x + 4) runs from 2 to 4, so the area must lie between 8 and 16, which rules out 56/9 and 64/9.

Answer: (b) \( \frac{15}{4} \) sq. units

Explanation:
1. y = x³ is positive for 1 ≤ x ≤ 2.
2. Area \( = \int_{1}^{2} x^3\,dx = \left[\frac{x^4}{4}\right]_{1}^{2} = \frac{16 - 1}{4} \).
3. \( = \frac{15}{4} \) sq. units.

Teacher's Note:
1. Check the sign of the curve on the interval before integrating; here it is positive throughout.
2. A common slip is to forget to subtract the value at the lower limit.

Answer: (b) 1

Explanation:
1. Area \( = \int_{0}^{2} 2^{kx}\,dx = \left[\frac{2^{kx}}{k\ln 2}\right]_{0}^{2} = \frac{2^{2k} - 1}{k\ln 2} \).
2. Given this equals \( \frac{3}{\ln 2} \): \( \frac{4^k - 1}{k} = 3 \).
3. k = 1 gives (4 − 1)/1 = 3, so k = 1.

Teacher's Note:
1. Remember \( \int a^{x}dx = \frac{a^x}{\ln a} \); the ln 2 in the answer is a hint that this formula is needed.
2. Testing the options in the simplified equation is faster than solving it algebraically.

Answer: (d) \( \frac{255}{4} \) sq. units

Explanation:
1. y = x³ is positive for 1 ≤ x ≤ 4.
2. Area \( = \int_{1}^{4} x^3\,dx = \left[\frac{x^4}{4}\right]_{1}^{4} = \frac{256 - 1}{4} \).
3. \( = \frac{255}{4} \) sq. units.

Teacher's Note:
1. 4⁴ = 256 is worth remembering; many students write 64 by mistake.
2. Options 64 and 127/4 come from using x³ instead of x⁴/4, so check your antiderivative.

Answer: (c) 2c log 2 sq. units

Explanation:
1. y = c/x, which is positive for x > 0 (taking c > 0).
2. Area \( = \int_{1}^{4} \frac{c}{x}\,dx = c[\log x]_{1}^{4} = c\log 4 \).
3. c log 4 = 2c log 2 sq. units.

Teacher's Note:
1. log 4 = 2 log 2 converts the answer into the form given in the options.
2. The rectangular hyperbola xy = c gives a logarithmic area, so the answer has no powers of x.

Answer: (a) 1

Explanation:
1. The curve y = logₑx cuts the x-axis at x = 1 and is below the axis for 0 < x < 1; the region with the axes is there.
2. \( \int_{0}^{1} \log_e x\,dx = [x\log_e x - x]_{0}^{1} = (0 - 1) - 0 = -1 \), using \( x\log x \to 0 \) as x → 0.
3. Area = |−1| = 1 sq. unit.

Teacher's Note:
1. \( \int \log x\,dx = x\log x - x \) comes from integration by parts with log x as the first function.
2. Though the curve goes to −∞ along the y-axis, the area is finite, so do not choose ∞.

Answer: (b) 1

Explanation:
1. y = log x meets the x-axis at x = 1 and is positive for x > 1, so the region is from x = 1 to x = e.
2. Area \( = \int_{1}^{e} \log x\,dx = [x\log x - x]_{1}^{e} = (e - e) - (0 - 1) \).
3. = 1 sq. unit.

Teacher's Note:
1. The lower limit is where the curve cuts the axis, x = 1, not x = 0.
2. log e = 1 and log 1 = 0; using these makes the evaluation quick.

Answer: (c) log 4 − 1 sq. units

Explanation:
1. log x ≥ 0 for 1 ≤ x ≤ 2.
2. Area \( = \int_{1}^{2} \log x\,dx = [x\log x - x]_{1}^{2} = (2\log 2 - 2) - (0 - 1) \).
3. = 2 log 2 − 1 = log 4 − 1 sq. units.

Teacher's Note:
1. Write 2 log 2 as log 4 to match the options.
2. Check: log 4 ≈ 1.386, so the area ≈ 0.386, a sensible value under a curve that rises from 0 to about 0.69.

Answer: (b) \( \frac{e^{a^2} - 1}{2} \) sq. units

Explanation:
1. Put t = x², dt = 2x dx; when x = 0, t = 0 and when x = a, t = a².
2. Area \( = \int_{0}^{a} xe^{x^2}dx = \frac{1}{2}\int_{0}^{a^2} e^{t}dt = \frac{1}{2}\left[e^{t}\right]_{0}^{a^2} \).
3. \( = \frac{e^{a^2} - 1}{2} \) sq. units.

Teacher's Note:
1. When the derivative of the power (2x) appears as a factor, substitute the power; this is the standard trick for x·e^(x²).
2. Check with a = 0: the area must be 0, which only the expressions with e^(a²) − 1 give.

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

Explanation:
1. y = x²/4 ≥ 0, and the region runs from x = 0 to x = 2.
2. Area \( = \int_{0}^{2} \frac{x^2}{4}\,dx = \left[\frac{x^3}{12}\right]_{0}^{2} \).
3. \( = \frac{8}{12} = \frac{2}{3} \) sq. units.

Teacher's Note:
1. The region between a parabola, the x-axis and a line x = c is one-third of the enclosing rectangle: here \( \frac{1}{3} \times 2 \times 1 = \frac{2}{3} \).
2. Write y in terms of x before integrating with respect to x.

Answer: (c) \( \frac{7}{3} \) sq. units

Explanation:
1. The region is bounded by the y-axis, so integrate with respect to y: x = \( \frac{\sqrt{y}}{2} \).
2. Area \( = \int_{1}^{4} \frac{\sqrt{y}}{2}\,dy = \frac{1}{2}\cdot\frac{2}{3}\left[y^{3/2}\right]_{1}^{4} = \frac{1}{3}(8 - 1) \).
3. \( = \frac{7}{3} \) sq. units.

Teacher's Note:
1. When the boundaries are horizontal lines y = c and the y-axis, use horizontal strips and integrate x dy.
2. Take the positive root x = √y/2 for the part on the right of the y-axis.

Answer: (a) \( 4a^2 \) sq. units

Explanation:
1. y = 0 gives x = ±a, and y > 0 between them (taking a > 0).
2. Area \( = \int_{-a}^{a} \frac{3}{a}(a^2 - x^2)\,dx = \frac{6}{a}\int_{0}^{a}(a^2 - x^2)\,dx = \frac{6}{a}\left(a^3 - \frac{a^3}{3}\right) \).
3. \( = \frac{6}{a}\cdot\frac{2a^3}{3} = 4a^2 \) sq. units.

Teacher's Note:
1. The curve is symmetric about the y-axis, so compute from 0 to a and double it.
2. Dimension check: an area must be in units of a², which rules out 4a³.

Answer: (c) 1

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

Teacher's Note:
1. One arch of y = sin x (from 0 to π) has area 2, so a half-arch has area 1.
2. Do not confuse the length of the interval (π/2) with the area.

Answer: (b) 4 sq. units

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

Teacher's Note:
1. Integrating straight from 0 to 2π gives 0 because the two parts cancel; split the interval at every zero of the curve.
2. Each arch of sin x or cos x has area 2, so n arches give 2n.

Answer: (a) 2k sq. units

Explanation:
1. On [π, 2π], sin x ≤ 0, so k sin x ≤ 0 for k > 0 and the region is below the axis.
2. \( \int_{\pi}^{2\pi} k\sin x\,dx = k[-\cos x]_{\pi}^{2\pi} = k(-1 - 1) = -2k \).
3. Area = |−2k| = 2k sq. units.

Teacher's Note:
1. The integral being 0 is a trap only when the interval covers both signs; here the sign is constant, so just take the modulus.
2. This is one arch of sin x stretched k times vertically, so the area is k × 2.

Answer: (c) 2 log 2

Explanation:
1. tan x < 0 on (−π/3, 0) and > 0 on (0, π/3); the two parts are mirror images.
2. Area \( = 2\int_{0}^{\pi/3}\tan x\,dx = 2[\log\sec x]_{0}^{\pi/3} = 2(\log 2 - \log 1) \).
3. = 2 log 2 sq. units.

Teacher's Note:
1. tan x is an odd function, so its integral from −π/3 to π/3 is 0; option (d) is the integral, not the area.
2. \( \int \tan x\,dx = \log|\sec x| + C \), and sec(π/3) = 2.

Answer: (b) \( \frac{\pi}{4} \)

Explanation:
1. \( \sin^2 x = \frac{1 - \cos 2x}{2} \ge 0 \).
2. Area \( = \int_{0}^{\pi/2}\frac{1 - \cos 2x}{2}\,dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_{0}^{\pi/2} \).
3. \( = \frac{1}{2}\cdot\frac{\pi}{2} = \frac{\pi}{4} \) sq. units.

Teacher's Note:
1. Shortcut: \( \int_{0}^{\pi/2}\sin^2 x\,dx = \int_{0}^{\pi/2}\cos^2 x\,dx \), and the two add to π/2, so each is π/4.
2. Always reduce sin²x with the double-angle formula before integrating.

Answer: (d) \( 2\sqrt{3} : (4 - \sqrt{3}) \)

Explanation:
1. Area under y = cos x: \( \int_{0}^{\pi/3}\cos x\,dx = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \).
2. cos 2x ≥ 0 on [0, π/4] and < 0 on (π/4, π/3]: \( \int_{0}^{\pi/4}\cos 2x\,dx = \frac{1}{2} \), \( \int_{\pi/4}^{\pi/3}\cos 2x\,dx = \frac{\sqrt{3}}{4} - \frac{1}{2} \).
3. Area under cos 2x \( = \frac{1}{2} + \left(\frac{1}{2} - \frac{\sqrt{3}}{4}\right) = \frac{4 - \sqrt{3}}{4} \).
4. Ratio \( = \frac{\sqrt{3}}{2} : \frac{4 - \sqrt{3}}{4} = 2\sqrt{3} : (4 - \sqrt{3}) \).

Teacher's Note:
1. cos 2x changes sign at x = π/4, inside the interval; ignoring this gives the wrong value \( \frac{\sqrt{3}}{4} \) for its area.
2. Multiply both parts of a ratio by the same number (here 4) to clear fractions.

Answer: (a) \( \log(\sqrt{2} + 1) \)

Explanation:
1. sec x > 0 on [0, π/4].
2. Area \( = \int_{0}^{\pi/4}\sec x\,dx = [\log(\sec x + \tan x)]_{0}^{\pi/4} \).
3. \( = \log(\sqrt{2} + 1) - \log 1 = \log(\sqrt{2} + 1) \) sq. units.

Teacher's Note:
1. Learn \( \int \sec x\,dx = \log|\sec x + \tan x| + C \); it is used often in area and definite integral questions.
2. log(√2 − 1) is negative, so it can never be an area.

Answer: (d) \( \frac{7}{10} \)

Explanation:
1. For 1 ≤ x < 2, [x] = 1, so the curve is the horizontal line y = 1 on the interval.
2. Area \( = \int_{1}^{1.7} 1\,dx = 1.7 - 1 \).
3. \( = 0.7 = \frac{7}{10} \).

Teacher's Note:
1. For greatest-integer functions, split the interval at integers; on each piece [x] is a constant.
2. The region here is just a rectangle of width 0.7 and height 1.

Answer: (b) \( \sqrt[3]{8 - \sqrt{17}} \)

Explanation:
1. y = x²(8 − x³) ≥ 0 for x ≤ 2, so for 1 < k ≤ 2: \( \int_{1}^{k}(8x^2 - x^5)\,dx = \frac{8k^3}{3} - \frac{k^6}{6} - \frac{5}{2} \).
2. Setting this equal to \( \frac{16}{3} \): \( \frac{16k^3 - k^6}{6} = \frac{47}{6} \), i.e., \( k^6 - 16k^3 + 47 = 0 \).
3. So \( k^3 = 8 \pm \sqrt{17} \); only \( 8 - \sqrt{17} \) (≈ 3.88) keeps k below 2, where the curve is above the axis.
4. Hence \( k = \sqrt[3]{8 - \sqrt{17}} \).

Teacher's Note:
1. Treat the equation as a quadratic in k³; substituting u = k³ makes it routine.
2. Reject a root that takes the region across a zero of the curve, since the integral then no longer equals the area.