JEE Mathematics Definite Integration MCQs Set 22

Answer: (b) \( a-2a\log2 \)

Explanation:
1. (x - a)/(x + a) = 1 - 2a/(x + a).
2. ∫0^a = [x - 2a log(x + a)] = a - 2a(log 2a - log a) = a - 2a 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.

Answer: (c) \( g(x) = -g(a - x) \)

Explanation:
1. Replacing x by a - x gives I = ∫0^a g(a - x)/(f(a - x) + f(x)) dx.
2. Adding, 2I = ∫0^a [g(x) + g(a - x)]/(f(x) + f(a - x)) dx.
3. This is 0 if g(x) + g(a - x) = 0, i.e. g(x) = -g(a - x).

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.

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

Explanation:
1. √(1 + sin 2θ) = |sin θ + cos θ|.
2. sin θ + cos θ > 0 on (0, 3π/4) and < 0 on (3π/4, π).
3. The integrand is +1 and then -1: 3π/4 - π/4 = 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.

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

Explanation:
1. Put x = a sin θ: ∫0^(π/2) cos θ dθ/(sin θ + cos θ).
2. By θ → π/2 - θ the pair adds to 1, so I = 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.

Answer: (c) \( \pi \)

Explanation:
1. The integrand is continuous and has period π, so I = 2∫0^π = 4∫0^(π/2) (using symmetry).
2. By x → π/2 - x, 1/(1 + tan⁴x) + 1/(1 + cot⁴x) = 1, so ∫0^(π/2) = π/4.
3. I = 4(π/4) = π.

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.

Answer: (a) \( -1 \)

Explanation:
1. d/dx(e^x cos x) = e^x(cos x - sin x).
2. So the integral is [e^x cos x] from 0 to π/2 = 0 - 1 = -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.

Answer: (d) \( - \frac{3}{4} + \log{\left(4 \right)} \)

Explanation:
1. By parts: ∫(1 + x) log(1 + x) dx = ((1 + x)²/2) log(1 + x) - ∫(1 + x)/2 dx.
2. From 0 to 1: 2 log 2 - (1/4)(4 - 1) = 2 log 2 - 3/4 = log 4 - 3/4.

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.

Answer: (d) \( for all values of a and b \)

Explanation:
1. Both x³ and x are odd functions.
2. So ax³ + bx is odd for every a and b, and the integral over [-1, 1] 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.

Answer: (a) \( 26 \)

Explanation:
1. [2x + 3] = 3 + [2x].
2. ∫0^4 3 dx = 12. [2x] takes the values 0, 1, ..., 7 on intervals of length 1/2, so ∫0^4 [2x] dx = (1/2)(0 + 1 + ... + 7) = 14.
3. Total = 26.

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.

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

Explanation:
1. Put x = tan θ: the integral becomes ∫0^(π/4) cos θ dθ.
2. = sin(π/4) = √2/2.

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.

Answer: (c) \( 3 \)

Explanation:
1. By parts: I_n = -(-1)^n·... = (-1)^(n+1) - n I_(n-1) with I₀ = e - 1.
2. I₁ = 2 - e, I₂ = 2e - 5, I₃ = 16 - 6e.
3. So n = 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.

Answer: (b) \( 5051 I₁ = 5050 I₂ \)

Explanation:
1. Integrating I₁ by parts: I₁ = [x(1 - x⁵⁰)^101] - ∫x · 101(1 - x⁵⁰)^100(-50x⁴⁹) dx = 5050∫x⁵⁰(1 - x⁵⁰)^100 dx.
2. Now x⁵⁰ = 1 - (1 - x⁵⁰), so ∫x⁵⁰(1 - x⁵⁰)^100 dx = I₂ - I₁.
3. I₁ = 5050(I₂ - I₁), hence 5051 I₁ = 5050 I₂.

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.

Answer: (c) \( \frac a2 \)

Explanation:
1. Let I be the integral. Replacing x by a - x gives I = ∫0^a f(a - x)/(f(a - x) + f(x)) dx.
2. Adding, 2I = ∫0^a 1 dx = a.
3. So I = a/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.

Answer: (d) \( \frac n{\log2} \)

Explanation:
1. 2^x/2^[x] = 2^(x - [x]) = 2^({x}), which has period 1.
2. ∫0^n = n∫0^1 2^x dx = n/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.

Answer: (c) \( \tan^{-1}{\left(4 \right)} \)

Explanation:
1. ∫dx/(1 + x²) = tan⁻¹x.
2. From 0 to 4: tan⁻¹4.

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.

Answer: (c) \( \frac\pi2-\tan^{-1}\frac2{\sqrt3} \)

Explanation:
1. [√3 tan x] = 0 for x < π/6, 1 for π/6 ≤ x < tan⁻¹(2/√3), and 2 for tan⁻¹(2/√3) ≤ x < π/3.
2. ∫ = (tan⁻¹(2/√3) - π/6) + 2(π/3 - tan⁻¹(2/√3)).
3. = π/2 - tan⁻¹(2/√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.

Answer: (b) \( \frac a{a^2+b^2}\left(e^{2a\pi}-1\right) \)

Explanation:
1. ∫e^(ax) cos bx dx = e^(ax)(a cos bx + b sin bx)/(a² + b²).
2. At 2π with integer b: cos 2πb = 1, sin 2πb = 0, giving a e^(2πa)/(a² + b²).
3. At 0: a/(a² + b²). The difference is a(e^(2aπ) - 1)/(a² + b²).

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.

Answer: (a) \( \frac{5}{24} \)

Explanation:
1. Put x + √(x² + 1) = t, so x = (t - 1/t)/2 and dx = (1/2)(1 + 1/t²) dt; limits t = 1 to ∞.
2. ∫ = (1/2)∫1^∞ (t^(-5) + t^(-7)) dt = (1/2)(1/4 + 1/6) = 5/24.

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.

Answer: (a) \( A=e,\ B=-n \)

Explanation:
1. By parts: I_n = [x (log x)^n] from 1 to e - n∫1^e (log x)^(n-1) dx.
2. = e - n I_(n-1).
3. So A = e and B = -n.

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.

Answer: (c) \( 2e^{(\pi-4)/2} \)

Explanation:
1. log y = lim (1/n)Σ log(1 + r²/n²) = ∫0^1 log(1 + x²) dx.
2. By parts: [x log(1 + x²)] - ∫2x²/(1 + x²) dx = log 2 - 2 + 2·(π/4) = log 2 - 2 + π/2.
3. y = e^(log 2 - 2 + π/2) = 2e^((π - 4)/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.

Answer: (b) \( \frac12\tan1 \)

Explanation:
1. The r-th term is (1/n)(r/n) sec²((r/n)²).
2. The limit is ∫0^1 x sec²(x²) dx.
3. Put x² = t: (1/2)∫0^1 sec²t dt = (1/2) tan 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.

Answer: (d) \( 0 \)

Explanation:
1. sin mx sin nx = (1/2)[cos(m - n)x - cos(m + n)x].
2. Since m - n and m + n are non-zero integers, both terms integrate to 0 over [0, π].

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.