JEE Mathematics Definite Integration MCQs Set 10

Answer: (a) \( \sqrt e \)

Explanation:
1. For 1 ≤ x < a, log_a x ∈ [0, 1), so [log_a x] = 0 and the integrand is x.
2. \( \int_1^ax\,dx=\frac{a^2-1}{2}=\frac{e-1}{2} \), so a² = e and a = √e.

Teacher's Note:
1. On the whole interval [1, a) the greatest integer is 0.
2. A single point (x = a) does not affect the integral.

Answer: (d) t ∈ (0, b] only

Explanation:
1. f(x) > 0 for every x in [a, b].
2. For t > 0, F(t) is the area under the curve from 0 to t, which is positive.
3. For t < 0, F(t) = −∫ₜ⁰ f(x)dx < 0, and F(0) = 0.
4. So F(t) > 0 only for t ∈ (0, b].

Teacher's Note:
1. The sign of ∫₀ᵗ f depends on the direction of integration as well as the sign of f.
2. Integrating backwards (t < 0) reverses the sign.

Answer: (d) \( \begin{cases}x & x\le0\\ e^x-1 & x>0\end{cases} \)

Explanation:
1. Put u = ln x. For 0 < x ≤ 1, u ≤ 0 and f′(u) = 1; for x > 1, u > 0 and f′(u) = eᵘ.
2. Integrating with f(0) = 0: f(u) = u for u ≤ 0 and f(u) = eᵘ − 1 for u > 0 (continuous at 0).

Teacher's Note:
1. Change the variable to u = ln x so that f′(u) is given in terms of u.
2. Use continuity at 0 and f(0) = 0 to fix the constants.

Answer: (c) \( \frac{\tan^{-1}x}{x} \)

Explanation:
1. Write the term as \( \frac1n\cdot\frac1{1+(k/n)^2x^2} \).
2. \( \int_0^1\frac{dt}{1+x^2t^2}=\frac{\tan^{-1}x}{x} \).

Teacher's Note:
1. Factor n² from the denominator to recognise a Riemann sum.
2. Substitute t = k/n.

Answer: (b) 8

Explanation:
1. The sum is the right-hand Riemann sum of f(x) = x on [0, 4].
2. \( \int_0^4x\,dx=8 \).

Teacher's Note:
1. A sum of xᵢ δx as δx → 0 is the definite integral of x.
2. The limits are the end points of the interval, 0 and 4.

Answer: (a) J + K = 0

Explanation:
1. In J, put x = 1/u: \( J=\int_0^\infty\frac{-\ln u}{1+u^{-3}}\cdot\frac{du}{u^2}=-\int_0^\infty\frac{u\ln u}{u^3+1}du=-K \).
2. Hence J + K = 0.

Teacher's Note:
1. The substitution x → 1/x maps J onto the negative of K.
2. Check the algebra: u^{-2}/(1 + u^{-3}) = u/(u³ + 1).

Answer: (c) e² − 1

Explanation:
1. As n → ∞, \( \left(1+\frac t{n+1}\right)^n\to e^t \) uniformly on [0, 2].
2. \( \int_0^2e^tdt=e^2-1 \).

Teacher's Note:
1. The integrand converges to eᵗ uniformly, so the limit and the integral can be interchanged.
2. Use (1 + t/n)ⁿ → eᵗ.

Answer: (a) e

Explanation:
1. \( \ln\frac{(n!)^{1/n}}{n}=\frac1n\sum_{k=1}^n\ln\frac kn\to\int_0^1\ln x\,dx=-1 \).
2. So (n!)^{1/n}/n → e⁻¹ and the required limit is e.

Teacher's Note:
1. Take logarithms to turn the product into a Riemann sum.
2. ∫₀¹ ln x dx = −1.

Answer: (a) sin(cosec 1) − sin(cosec 2)

Explanation:
1. Put u = 1/sin x, so cos x dx/sin²x = −du; the limits become u = 2 (x = π/6) to u = 1 (x = π/2).
2. Put v = 1/sin u, so cos u du/sin²u = −dv.
3. The integral becomes \( \int\cos v\,dv=\sin v \) with v = cosec 2 at the lower limit and v = cosec 1 at the upper limit.
4. Value: sin(cosec 1) − sin(cosec 2).

Teacher's Note:
1. Two nested substitutions u = 1/sin x and v = 1/sin u remove the complicated factors.
2. Track the limits carefully through both substitutions.

Answer: (b) 1/2

Explanation:
1. The weight n xⁿ⁻¹ concentrates near x = 1 and integrates to 1.
2. So the limit equals the value of 1/(1 + x) at x = 1, which is 1/2. (By parts: \( \left[\frac{x^n}{1+x}\right]_0^1+\int_0^1\frac{x^n}{(1+x)^2}dx\to\frac12 \).)

Teacher's Note:
1. Integrate by parts to see the limit exactly.
2. The second term vanishes as n → ∞.

Answer: (b) 6

Explanation:
1. The interval has length 18, which is 18 whole periods of both {x}² and sin 2πx.
2. Each period contributes ∫₀¹ x² dx = 1/3 and 0 respectively.
3. Total = 18 × 1/3 = 6.

Teacher's Note:
1. Integrate over complete periods; the periodic function repeats.
2. The sine part integrates to zero over each period.

Answer: (d) 2/3

Explanation:
1. For small x, \( \int_0^x\sin t^2dt\approx\frac{x^3}3 \) and \( x(1-\cos x)\approx\frac{x^3}2 \).
2. The ratio tends to (1/3)/(1/2) = 2/3.

Teacher's Note:
1. Use leading terms of the Taylor expansions near 0.
2. L'Hôpital's rule gives the same result.

Answer: (a) equal to zero

Explanation:
1. Put x = 1/u: \( I=\int_0^\infty f\left(u+\frac1u\right)\frac{-\ln u}{u}du=-I \).
2. So 2I = 0 and I = 0.

Teacher's Note:
1. The function x + 1/x is unchanged under x → 1/x while ln x changes sign.
2. An integral equal to its own negative is zero.

Answer: (b) \( \frac\pi{\sqrt2} \)

Explanation:
1. Use the standard result \( \int_0^{\pi/2}\tan^{k}x\,dx=\frac{\pi}{2\cos(k\pi/2)} \) for |k| < 1.
2. With k = 1/2: \( \frac{\pi}{2\cos(\pi/4)}=\frac{\pi}{\sqrt2} \).

Teacher's Note:
1. Substituting t = √tan x leads to ∫ 2t²/(1 + t⁴) dt, which is a standard integral.
2. The result for tan^k x is π/(2 cos(kπ/2)).

Answer: (a) \( \frac\pi4-\frac12 \)

Explanation:
1. Let g(x) = x tan⁻¹x (even). Replacing x by −x: \( I=\int\frac{g(x)\,e^{\tan^{-1}x}}{1+e^{\tan^{-1}x}}dx \). Adding: \( 2I=\int_{-1}^1g(x)dx \), so \( I=\int_0^1x\tan^{-1}x\,dx \).
2. By parts: \( \frac{\pi}{8}-\frac12\int_0^1\frac{x^2}{1+x^2}dx=\frac\pi8-\frac12\left(1-\frac\pi4\right)=\frac\pi4-\frac12 \).

Teacher's Note:
1. The weights 1/(1 + e^{tan⁻¹x}) and e^{tan⁻¹x}/(1 + e^{tan⁻¹x}) add up to 1.
2. Then integrate x tan⁻¹x by parts.

Answer: (a) \( \tan^2\left(\frac\pi8\right) \)

Explanation:
1. \( \int\frac{dx}{\sqrt x(\sqrt x+1)}=2\ln(1+\sqrt x) \), so the integral is \( 2\ln\frac{1+a}{1+\sqrt a} \).
2. Let s = √a and minimise (1 + s²)/(1 + s): the derivative is zero when s² + 2s − 1 = 0, i.e. s = √2 − 1 = tan(π/8).
3. So a = s² = tan²(π/8).

Teacher's Note:
1. Write x + √x = √x(√x + 1) to integrate in closed form.
2. Minimise the argument of the logarithm by differentiating.

Answer: (d) \( 1+2e^{-1} \)

Explanation:
1. \( f'(x)=1+\ln^2x+2\ln x=(1+\ln x)^2 \), which vanishes at x = 1/e.
2. An antiderivative of ln²t + 2 ln t is t ln²t, so \( \int_1^{1/e}=\frac1e \).
3. f(1/e) = 1 + 1/e + 1/e = 1 + 2e⁻¹.

Teacher's Note:
1. Differentiate to find where f′ vanishes, then evaluate f there.
2. d/dt (t ln²t) = ln²t + 2 ln t.

Answer: (b) \( \frac4e \)

Explanation:
1. \( \int_0^1(1+x)^\lambda dx=\frac{2^{\lambda+1}-1}{\lambda+1}\approx1+\lambda(2\ln2-1) \) for small λ.
2. So the limit is \( e^{2\ln2-1}=\frac4e \).

Teacher's Note:
1. Evaluate the integral exactly and expand for small λ (a 1^∞ form).
2. Use (1 + λc)^{1/λ} → e^c.

Answer: (d) (−∞, −1] ∪ [0, ∞)

Explanation:
1. With y = 3⁻ᵃ the integral equals \( \frac1{\ln3}\left(\frac32+\frac{y^2}2-2y\right) \).
2. The inequality is (y − 1)(y − 3) ≥ 0, so y ≤ 1 or y ≥ 3.
3. 3⁻ᵃ ≤ 1 gives a ≥ 0, and 3⁻ᵃ ≥ 3 gives a ≤ −1.

Teacher's Note:
1. Find the antiderivative and evaluate between a and 0.
2. Convert the condition in y back to a.

Answer: (b) 2

Explanation:
1. The integral is sin(α² + α) − sin α², so the equation becomes \( h(\alpha)=\sin(\alpha^2+\alpha)-\sin\alpha^2-\sin\alpha=0 \).
2. Evaluating h on (2, 3) shows two sign changes (h changes sign twice), so there are 2 solutions.

Teacher's Note:
1. Reduce the equation to the zeros of one function of α.
2. Check the sign of h at several points in (2, 3).

Answer: (b) \( 2e^4-e-\alpha \)

Explanation:
1. Put x = e^{u²}: dx = 2u e^{u²}du, √(ln x) = u, and u runs from 1 to 2.
2. \( \int_1^22u^2e^{u^2}du=\left[ue^{u^2}\right]_1^2-\int_1^2e^{u^2}du=2e^4-e-\alpha \).

Teacher's Note:
1. Substitute so that √(ln x) becomes a simple variable.
2. Integrate by parts with u and 2u e^{u²}.

Answer: (b) \( A-\frac e2+1 \)

Explanation:
1. By parts with u = eᵗ and dv = dt/(1 + t)²: \( \int_0^1\frac{e^t}{(1+t)^2}dt=\left[-\frac{e^t}{1+t}\right]_0^1+\int_0^1\frac{e^t}{1+t}dt \).
2. = −e/2 + 1 + A.

Teacher's Note:
1. Integrate 1/(1 + t)² to get −1/(1 + t) and move the derivative onto eᵗ.
2. The remaining integral is exactly A.

Answer: (a) \( e^{-1} \)

Explanation:
1. By parts: \( \int_0^1e^{-x^2}dx=\left[xe^{-x^2}\right]_0^1+2\int_0^1x^2e^{-x^2}dx=e^{-1}+2\int_0^1x^2e^{-x^2}dx \).
2. Comparing with the given relation, β = e⁻¹.

Teacher's Note:
1. Integrate e^{−x²} by parts taking x as the derivative factor.
2. The new integral has the same form as the one in the question.