JEE Mathematics Definite Integration MCQs Set 24

Multiple Choice Questions (MCQs) for JEE Mathematics: Definite Integration

Review structured MCQ sets for JEE Mathematics Definite Integration. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Access the complete set of multiple-choice questions for Definite Integration below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: \( \int\limits_{0}^{2 \pi} x \sin^{4}{\left(x \right)} \cos^{6}{\left(x \right)}\, dx = \)
(a) \( \frac{3 \pi^{2}}{64} \)
(b) \( \frac{5 \pi^{2}}{128} \)
(c) \( \frac{15 \pi^{2}}{128} \)
(d) \( \frac{3 \pi^{2}}{128} \)
Show Answer & Explanation

Answer: (d) \( \frac{3 \pi^{2}}{128} \)

Explanation:
1. I = π∫0^(2π) sin⁴x cos⁶x dx (by x → 2π - x), = 4π∫0^(π/2) sin⁴x cos⁶x dx.
2. By Wallis, ∫0^(π/2) sin⁴x cos⁶x dx = (3·1)(5·3·1)/(10·8·6·4·2) × π/2 = 3π/512.
3. I = 4π × 3π/512 = 3π²/128.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 2: Evaluate \( \int_0^1\cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right)dx \) =
(a) \( -\frac12 \)
(b) \( 1 \)
(c) \( \frac12 \)
(d) \( 0 \)
Show Answer & Explanation

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

Explanation:
1. Put x = cos 2θ: √((1 - x)/(1 + x)) = tan θ, so cot⁻¹ tan θ = π/2 - θ.
2. The integrand is cos(π - 2θ) = -cos 2θ = -x.
3. ∫0^1 -x dx = -1/2.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 3: If \( f(x)=\begin{cases}e^{\cos x}\sin x,&|x|\le2\\2,&\text{otherwise}\end{cases} \), then \( \int_{-2}^{3}f(x)\,dx \) =
(a) \( 0 \)
(b) \( 2 \)
(c) \( 1 \)
(d) \( 3 \)
Show Answer & Explanation

Answer: (b) \( 2 \)

Explanation:
1. On [-2, 2] the function e^(cos x) sin x is odd, so its integral is 0.
2. On (2, 3] the function is 2, giving 2(3 - 2) = 2.
3. Total = 2.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 4: Evaluate \( \int_0^{\pi/2}\frac{a\cos x+b\sin x}{\cos x+\sin x}dx \) =
(a) \( \frac\pi4(a+b) \)
(b) \( \pi(a+b) \)
(c) \( \pi ab \)
(d) \( \frac\pi2(a+b) \)
Show Answer & Explanation

Answer: (a) \( \frac\pi4(a+b) \)

Explanation:
1. By the property f(x) → f(π/2 - x), the integral also equals ∫(a sin x + b cos x)/(sin x + cos x) dx.
2. Adding, 2I = (a + b) ∫0^(π/2) 1 dx = (a + b)π/2.
3. So I = π(a + b)/4.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 5: Evaluate \( \int_{-1}^{1}\frac{d}{dx}\left(\tan^{-1}\frac1x\right)dx \) =
(a) \( -\frac\pi2 \)
(b) \( -\pi \)
(c) \( \frac\pi2 \)
(d) \( \pi \)
Show Answer & Explanation

Answer: (a) \( -\frac\pi2 \)

Explanation:
1. d/dx tan⁻¹(1/x) = -1/(1 + x²) for x ≠ 0.
2. ∫_(-1)^1 -dx/(1 + x²) = -[tan⁻¹x] = -(π/4 + π/4) = -π/2.
3. (Do not apply F(1) - F(-1) directly: tan⁻¹(1/x) is discontinuous at x = 0.)

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 6: If \( \int_0^1\frac{\tan^{-1}x}{x}dx=k\int_0^{\pi/2}\frac{x}{\sin x}dx \), then \( k \) is
(a) \( 1 \)
(b) \( 4 \)
(c) \( \frac14 \)
(d) \( \frac12 \)
Show Answer & Explanation

Answer: (d) \( \frac12 \)

Explanation:
1. Put x = tan θ: ∫0^1 tan⁻¹x/x dx = ∫0^(π/4) θ sec²θ/tan θ dθ = ∫0^(π/4) 2θ/sin 2θ dθ.
2. Put 2θ = t: this is (1/2)∫0^(π/2) t/sin t dt.
3. Hence k = 1/2.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 7: \( \int\limits_{1}^{4} \frac{x}{\sqrt{4 x + 2}}\, dx = \)
(a) \( \frac{1}{2} \)
(b) \( \frac{3}{2} \)
(c) \( \frac{3 \sqrt{2}}{2} \)
(d) \( \frac{\sqrt{2}}{2} \)
Show Answer & Explanation

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

Explanation:
1. Put 2 + 4x = t², so x = (t² - 2)/4 and dx = t dt/2.
2. The integral is (1/8)∫_(√6)^(√18) (t² - 2) dt = (1/8)[t³/3 - 2t].
3. Evaluating gives 3/√2.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 8: For \( a>0 \), \( \int_0^\infty e^{-ax}\,dx \) =
(a) \( \frac1a \)
(b) \( 0 \)
(c) \( -\frac1a \)
(d) \( a \)
Show Answer & Explanation

Answer: (a) \( \frac1a \)

Explanation:
1. ∫0^∞ e^(-ax) dx = [-e^(-ax)/a] from 0 to ∞.
2. = 0 + 1/a = 1/a.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 9: \( \int\limits_{1}^{\sqrt[7]{2}} \frac{1}{x \left(2 x^{7} + 1\right)}\, dx = \)
(a) \( \frac{\log{\left(\frac{6}{5} \right)}}{7} \)
(b) \( 6 \log{\left(\frac{6}{5} \right)} \)
(c) \( \frac{\log{\left(\frac{6}{5} \right)}}{12} \)
(d) \( \log{\left(\frac{6}{5} \right)} \)
Show Answer & Explanation

Answer: (a) \( \frac{\log{\left(\frac{6}{5} \right)}}{7} \)

Explanation:
1. Put x⁷ = t, so 7x⁶ dx = dt and dx/x = dt/(7t).
2. ∫1^2 dt/(7t(2t + 1)) = (1/7)[log t - log(2t + 1)] from 1 to 2.
3. = (1/7)(log(2/5) - log(1/3)) = (1/7) log(6/5).

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 10: If \( f \) is a continuous periodic function on \( \mathbb R \) with period \( T \), then \( \int_{mT}^{nT}f(x)\,dx \) =
(a) \( \int_0^Tf(x)dx \)
(b) \( (n-m)\int_0^Tf(x)dx \)
(c) \( (n+m)\int_0^Tf(x)dx \)
(d) \( \frac{n+m}2\int_0^Tf(x)dx \)
Show Answer & Explanation

Answer: (b) \( (n-m)\int_0^Tf(x)dx \)

Explanation:
1. The interval [mT, nT] has length (n - m)T, which contains n - m full periods.
2. Each period gives the same integral ∫0^T f.
3. So the value is (n - m)∫0^T f(x) dx.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 11: The integral \( I=\int_0^1e^{x^2}dx \) satisfies the inequality
(a) \( -1\le I\le0 \)
(b) \( -1\le I\le1 \)
(c) \( 0(d) \( 1\le I\le e \)
Show Answer & Explanation

Answer: (d) \( 1\le I\le e \)

Explanation:
1. For 0 ≤ x ≤ 1, 0 ≤ x² ≤ 1, so 1 ≤ e^(x²) ≤ e.
2. Integrating over an interval of length 1: 1 ≤ I ≤ e.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 12: Evaluate \( \lim_{n\to\infty}\frac{1^m+2^m+3^m+\cdots+n^m}{n^{m+1}} \) =
(a) \( \frac1{m+3} \)
(b) \( \frac1m \)
(c) \( \frac1{m+2} \)
(d) \( \frac1{m+1} \)
Show Answer & Explanation

Answer: (d) \( \frac1{m+1} \)

Explanation:
1. Write the expression as (1/n) Σ (r/n)^m for r = 1 to n.
2. The limit is ∫0^1 x^m dx = 1/(m + 1).

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 13: Evaluate \( \lim_{n\to\infty}\frac{(n!)^{1/n}}n \) =
(a) \( e - 1 \)
(b) \( e - 5 \)
(c) \( \frac1e \)
(d) \( e \)
Show Answer & Explanation

Answer: (c) \( \frac1e \)

Explanation:
1. Let y be the limit; log y = lim (1/n) Σ log(r/n) = ∫0^1 log x dx = -1.
2. So y = 1/e.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 14: Evaluate \( \int_0^{2a}\frac{f(x)}{f(x)+f(2a-x)}dx \) =
(a) \( 0 \)
(b) \( 2a \)
(c) \( \frac a2 \)
(d) \( a \)
Show Answer & Explanation

Answer: (d) \( a \)

Explanation:
1. Replace x by 2a - x: I = ∫0^(2a) f(2a - x)/(f(2a - x) + f(x)) dx.
2. Adding, 2I = ∫0^(2a) dx = 2a.
3. So I = a.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 15: \( \int\limits_{0}^{\frac{\pi}{4}} \tan^{2}{\left(x \right)}\, dx = \)
(a) \( \frac{\pi}{2} \)
(b) \( \frac{\pi}{4} \)
(c) \( \frac{\pi}{4} + 1 \)
(d) \( 1 - \frac{\pi}{4} \)
Show Answer & Explanation

Answer: (d) \( 1 - \frac{\pi}{4} \)

Explanation:
1. tan²x = sec²x - 1.
2. ∫0^(π/4) = [tan x - x] = 1 - π/4.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 16: \( \int\limits_{0}^{1} \frac{\log{\left(x \right)}}{\sqrt{1 - x^{2}}}\, dx = \)
(a) \( 2^{- \log{\left(2 \right)}} \)
(b) \( - \pi \log{\left(2 \right)} \)
(c) \( - \frac{\pi \log{\left(2 \right)}}{2} \)
(d) \( \pi \log{\left(2 \right)} \)
Show Answer & Explanation

Answer: (c) \( - \frac{\pi \log{\left(2 \right)}}{2} \)

Explanation:
1. Put x = sin θ: the integral becomes ∫0^(π/2) log sin θ dθ.
2. This equals -(π/2) log 2.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 17: \( \int\limits_{0}^{\pi} x \log{\left(\sin{\left(x \right)} \right)}\, dx = \)
(a) \( \pi^{2} \log{\left(2 \right)} \)
(b) \( \frac{\pi^{2}}{4} \)
(c) \( - \frac{\pi^{2} \log{\left(2 \right)}}{2} \)
(d) \( \frac{\pi^{2} \log{\left(2 \right)}}{2} \)
Show Answer & Explanation

Answer: (c) \( - \frac{\pi^{2} \log{\left(2 \right)}}{2} \)

Explanation:
1. I = (π/2)∫0^π log sin x dx = π∫0^(π/2) log sin x dx.
2. = π(-(π/2) log 2) = -(π²/2) log 2.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 18: If \( f \) and \( g \) are continuous on \( [-a,a] \), then \( \int_{-a}^{a}\frac{g(x)-g(-x)}{f(x)+f(-x)}dx \) =
(a) \( 0 \)
(b) \( 6 \)
(c) \( 4 \)
(d) \( 3 \)
Show Answer & Explanation

Answer: (a) \( 0 \)

Explanation:
1. The numerator g(x) - g(-x) is odd and the denominator f(x) + f(-x) is even.
2. So the integrand is an odd function.
3. Its integral over [-a, a] is 0.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 19: \( \int\limits_{0}^{1} \sqrt{\frac{x}{1 - x^{3}}}\, dx = \)
(a) \( \frac{\pi}{4} \)
(b) \( \frac{\pi}{3} \)
(c) \( \frac{\pi}{2} \)
(d) \( \frac{\pi}{6} \)
Show Answer & Explanation

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

Explanation:
1. Put x^(3/2) = t, so (3/2)x^(1/2) dx = dt.
2. The integral becomes (2/3)∫0^1 dt/√(1 - t²) = (2/3)(π/2) = π/3.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 20: \( \int\limits_{1}^{3} \frac{1}{x^{2} - 2 x \left\lfloor{x}\right\rfloor + \left\lfloor{x}\right\rfloor^{2} + 1}\, dx = \)
(a) \( \frac{\pi}{2} \)
(b) \( 2 \pi \)
(c) \( 3 \pi \)
(d) \( \pi \)
Show Answer & Explanation

Answer: (a) \( \frac{\pi}{2} \)

Explanation:
1. The denominator is (x - [x])² + 1.
2. On [1, 2), [x] = 1: ∫1^2 dx/((x - 1)² + 1) = π/4. On [2, 3), [x] = 2: ∫2^3 dx/((x - 2)² + 1) = π/4.
3. Total = π/2.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 21: \( \int\limits_{0}^{\pi} \frac{x \sin{\left(x \right)}}{\cos^{2}{\left(x \right)} + 1}\, dx = \)
(a) \( \frac{\pi}{3} \)
(b) \( \frac{\pi^{2}}{4} \)
(c) \( - \frac{\pi^{2}}{2} \)
(d) \( - \frac{\pi^{2}}{4} \)
Show Answer & Explanation

Answer: (b) \( \frac{\pi^{2}}{4} \)

Explanation:
1. By the property, I = (π/2)∫0^π sin x/(1 + cos²x) dx.
2. Put cos x = t: (π/2)∫_(-1)^1 dt/(1 + t²) = (π/2)(π/2) = π²/4.

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 22: \( \int\limits_{0}^{\frac{\pi}{2}} \frac{\left(\sin{\left(x \right)} + 1\right) e^{x}}{\cos{\left(x \right)} + 1}\, dx = \)
(a) \( e^{\frac{\pi}{2}} \)
(b) \( 0 \)
(c) \( \frac{\pi}{4} \)
(d) \( -1 + e^{\frac{\pi}{2}} \)
Show Answer & Explanation

Answer: (a) \( e^{\frac{\pi}{2}} \)

Explanation:
1. (1 + sin x)/(1 + cos x) = (1/2)sec²(x/2) + tan(x/2) = f + f' with f = tan(x/2).
2. So the integral is [e^x tan(x/2)] from 0 to π/2 = e^(π/2).

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

Question 23: \( \int\limits_{0}^{1} \frac{1}{x^{2} + 2 x \cos{\left(a \right)} + 1}\, dx = \)
(a) \( \frac{a}{2 \sin{\left(a \right)}} \)
(b) \( \tan^{-1}{\left(\sin{\left(a \right)} \right)} \)
(c) \( a \sin{\left(a \right)} \)
(d) \( \sin{\left(a \right)} \)
Show Answer & Explanation

Answer: (a) \( \frac{a}{2 \sin{\left(a \right)}} \)

Explanation:
1. x² + 2x cos α + 1 = (x + cos α)² + sin²α.
2. ∫0^1 = (1/sin α)[tan⁻¹((x + cos α)/sin α)] from 0 to 1.
3. The difference of angles is α/2, so I = α/(2 sin α).

Teacher's Note:
1. Look for a standard property of definite integrals such as symmetry, the king property or periodicity.
2. Split the interval where the integrand changes its form.

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