IIT JEE Mock Test – Integrals (Definite &

Indefinite) — Derived Level

Name: Integrals_mock_Derived_1

Pattern: 25 Questions | 10 Single-Correct MCQs | 5 Multiple-Correct MCQs | 5 Integer | 5 Numerical

Marking: +4 (correct MCQ), −1 (wrong MCQ); no negative for Integer/Numerical. Numerical answers
to 2 d.p.

Black & white • Font size 10 • Questions → Answer Key → Step-by-step Solutions
Question Paper

Section A — Single Correct MCQs (10)

1. Evaluate ∫₀¹ x²/(1+x²) dx. (A) 1−π/4 (B) π/4−1/2 (C) 1/2−π/4 (D) π/4

2. The value of ∫₀¹ ln x dx is: (A) −1 (B) 0 (C) 1 (D) −1/2

3. Evaluate ∫₀^{π/2} sin³x dx. (A) 1/2 (B) 2/3 (C) 3/4 (D) 1/3

4. The value of ∫₀^{π/2} ln(sin x) dx equals: (A) −π ln2 (B) −(π/2)ln2 (C) 0 (D) ln2

5. If F(x)=∫₀^x e^{t²} dt, then F′(x) equals: (A) 2x e^{x²} (B) e^{x²} (C) e^{2x} (D) cannot be
differentiated

6. Evaluate ∫₀¹ x/(1+x) dx. (A) 1−ln2 (B) ln2 (C) 1/2 (D) 1−1/2 ln2

7. If f is integrable on [0,a], then ∫₀^a f(x) dx equals: (A) ∫₀^a f(a−x) dx (B) −∫₀^a f(a−x) dx (C)
∫_{−a}^a f(x) dx (D) none

8. The improper integral ∫₁^∞ 1/x^p dx converges for: (A) p<1 (B) p=1 (C) p>1 (D) all p

9. Evaluate ∫₀^{π} |sin x| dx. (A) 1 (B) 2 (C) π (D) 4

10. An antiderivative of 1/[x(1+x)] is: (A) ln|x|+ln|1+x| (B) ln|x|−ln|1+x| (C) 1/(1+x)−1/x (D) −ln|
x|−ln|1+x|

Section B — Multiple Correct MCQs (5)

11. Let f be integrable on [−a,a]. Then: (A) If f is odd, ∫_{−a}^a f(x) dx=0 (B) If f is even, ∫_{−a}^a f(x)
dx=2∫₀^a f(x) dx (C) If f is odd, ∫₀^a f(x) dx=0 (D) If f is even, ∫₀^a f(x) dx=0

12. For I(m,n)=∫₀^{π/2} sin^m x cos^n x dx with m,n>−1: (A) I(m,n)=I(n,m) (B) I(m,n)= (m−1)/m ·
I(m−2,n) for m>0 (C) I(m,n)= (n−1)/n · I(m,n−2) for n>0 (D) I(2,2)=π/16

13. Regarding substitution in integrals: (A) For ∫ dx/(1+x²), x=tan θ gives ∫ dθ=θ+C (B) For
∫√(a²−x²) dx, x=a sin θ is useful (C) For ∫ dx/(x√(x²−a²)), x=a sec θ is useful (D) For ∫ e^{x} sin x dx,
use x=a tan θ

14. Convergence of improper integrals: (A) ∫₀¹ dx/√x converges (B) ∫₀¹ dx/x diverges (C) ∫₁^∞
dx/x² converges (D) ∫₀^∞ e^{−x} dx diverges

15. For J_n=∫₀^{π/2} sin^n x dx (n>0): (A) J_n = (n−1)/n · J_{n−2} (B) J_{2k} = ( (2k−1)!! / (2k)!! ) ·
(π/2) (C) J_{2k+1} = (2k)!!/(2k+1)!! (D) J_1=1

Section C — Integer-type (5)

16. Evaluate ∫₀^{π} sin x dx and enter the integer value.

17. Compute ⌊100 · ∫₀¹ dx/(1+x²)⌋ (greatest integer function).

18. Evaluate N = ∫₀^{π} |sin x| dx + ∫₀^{π} |cos x| dx (enter integer).

19. Evaluate ∫_{−1}^1 (x³ + x) dx (enter integer).

20. Evaluate ∫₀¹ ⌊3x⌋ dx (enter integer).

Section D — Numerical-type (5) (Answer to 2 d.p.)

21. Evaluate ∫₀¹ ln(1+x) dx (answer to 2 d.p.).

22. Evaluate ∫₀^{π/2} x sin x dx (answer to 2 d.p.).

23. Evaluate ∫₀¹ (x⁴ − x³ + x − 1/(1+x)) dx (answer to 2 d.p.).

24. Compute area under y=e^{−x} from x=0 to x=2 (i.e., ∫₀² e^{−x} dx) (to 2 d.p.).

25. Evaluate the improper integral ∫₀^{∞} e^{−1.25 x} dx (to 2 d.p.).

Answer Key — Summary

Section A — Single Correct MCQs

1-A, 2-A, 3-B, 4-B, 5-B, 6-A, 7-A, 8-C, 9-B, 10-B

Section B — Multiple Correct MCQs

11-A,B | 12-A,B,C,D | 13-A,B,C | 14-A,B,C | 15-A,B,C

Section C — Integer-type
16-2 | 17-78 | 18-4 | 19-0 | 20-1

Section D — Numerical-type
21-0.39 | 22-1.00 | 23--0.24 | 24-0.86 | 25-0.80
Step-by-Step Detailed Solutions

Section A — Single Correct MCQs

Q1. Solution:

• Rewrite integrand: x²/(1+x²) = 1 − 1/(1+x²).

• Integrate: ∫₀¹ (1 − 1/(1+x²)) dx = [x − arctan x]₀¹ = 1 − π/4.

• Correct option: (A) 1 − π/4.

Q2. Solution:

• Use ∫ ln x dx = x ln x − x.

• Evaluate from 0 to 1 using limit at 0⁺: (1·0 − 1) − (0 − 0) = −1.

• Correct: (A) −1.

Q3. Solution:

• Use reduction or identity: sin³x = sin x (1 − cos²x).

• Let u = cos x ⇒ du = −sin x dx. Then ∫₀^{π/2} sin³x dx = ∫₁^{0} (1−u²)(−du) = ∫₀^{1} (1−u²) du = [u
− u³/3]₀¹ = 2/3.

• Correct: (B) 2/3.

Q4. Solution:

• Classic result via Fourier/duplication/symmetry: ∫₀^{π/2} ln(sin x) dx = −(π/2) ln 2.

• Correct: (B).

Q5. Solution:

• By the Fundamental Theorem of Calculus: F′(x) = e^{x²}.

• Correct: (B).

Q6. Solution:

• Write x/(1+x) = 1 − 1/(1+x).

• ∫₀¹ x/(1+x) dx = [x − ln(1+x)]₀¹ = 1 − ln 2.

• Correct: (A).

Q7. Solution:

• Property: ∫₀^a f(x) dx = ∫₀^a f(a−x) dx (substitute x = a−t).

• Correct: (A).

Q8. Solution:

• p–test: ∫₁^∞ 1/x^p converges iff p>1.

• Correct: (C).

Q9. Solution:

• |sin x| on [0,π] equals sin x on [0,π]; ∫₀^{π} sin x dx = 2.

• Correct: (B).

Q10. Solution:

• Partial fractions: 1/[x(1+x)] = 1/x − 1/(1+x).

• Antiderivative: ln|x| − ln|1+x| + C.

• Correct: (B).

Section B — Multiple Correct MCQs

Q11. Solution:

• If f is odd on [−a,a], areas cancel ⇒ ∫_{−a}^a f=0 (A true).

• If f is even, integral doubles from 0..a ⇒ (B true).

• (C) is false in general; (D) false unless f=0.

Q12. Solution:

• Symmetry gives I(m,n)=I(n,m) (A).

• Reduction: I(m,n)= (m−1)/m I(m−2,n) for m>0 (B), similarly for n (C).

• Direct eval: I(2,2)=∫₀^{π/2} sin²x cos²x dx = (1/4)∫₀^{π/2} sin²2x dx = (1/8)∫₀^{π/2} (1−cos4x) dx

= (1/8)·(π/2) = π/16 (D).

• Thus A,B,C,D all correct.

Q13. Solution:

• (A) True: x=tanθ ⇒ dx=(1+tan²θ)dθ, integral simplifies to θ+C.

• (B) True: x=a sinθ simplifies √(a²−x²).

• (C) True: x=a secθ gives √(x²−a²)=a tanθ and dx fits.

• (D) False: for ∫ e^x sin x dx use integration by parts or complex exponentials, not trig substitution.

Q14. Solution:
• (A) ∫₀¹ x^{−1/2} dx converges (=2). (B) ∫₀¹ x^{−1} diverges (log).

• (C) ∫₁^∞ x^{−2} converges. (D) ∫₀^∞ e^{−x} dx converges (=1).

• So A,B,C true; D is false (it converges, statement says diverges).

Q15. Solution:

• Standard Wallis/reduction results: (A) J_n=(n−1)/n·J_{n−2}.

• (B) Even n=2k: J_{2k}=( (2k−1)!!/(2k)!! )·(π/2).

• (C) Odd n=2k+1: J_{2k+1}=(2k)!!/(2k+1)!!.

• (D) J_1=∫₀^{π/2} sin x dx = 1 ⇒ true.

• Thus A,B,C true; D true as well. (All A,B,C,D).

Section C — Integer-type
Q16. Solution:

• ∫ sin x dx = −cos x. Evaluate: [−cos x]₀^{π} = (−(−1)) − (−1) = 2.

Q17. Solution:

• ∫₀¹ dx/(1+x²) = [arctan x]₀¹ = π/4. Multiply by 100 ⇒ 25π ≈ 78.5398; floor ⇒ 78.

Q18. Solution:

• From earlier: ∫₀^{π} |sin x| dx = 2 and ∫₀^{π} |cos x| dx = 2.

• Sum N = 4.

Q19. Solution:

• Both x³ and x are odd; integral over [−1,1] cancels.

• Value = 0.

Q20. Solution:

• On [0,1): 3x∈[0,3). Floor is 0 on [0,1/3), 1 on [1/3,2/3), 2 on [2/3,1).

• Integral = 0·(1/3) + 1·(1/3) + 2·(1/3) = 1.

Section D — Numerical-type
Q21. Solution:

• Integrate by parts or known formula: ∫ ln(1+x) dx = (1+x)ln(1+x) − x.

• Evaluate 0→1: (2 ln2 − 1) − (0) = 2 ln2 − 1 ≈ 0.39.

Q22. Solution:
• Integrate by parts: let u=x, dv=sin x dx ⇒ du=dx, v=−cos x.

• ∫ x sin x dx = −x cos x + ∫ cos x dx = −x cos x + sin x.

• Evaluate 0→π/2: [−(π/2)·0 + 1] − [−0·1 + 0] = 1.00.

Q23. Solution:

• Integrate termwise: ∫₀¹ x⁴ dx=1/5; ∫ x³ dx=1/4; ∫ x dx=1/2; ∫ 1/(1+x) dx=ln(1+x).

• Total = (1/5) − (1/4) + (1/2) − ln2 = -0.24.

Q24. Solution:

• ∫ e^{−x} dx = −e^{−x}. Evaluate 0→2: (−e^{−2}) − (−1) = 1 − e^{−2} ≈ 0.86.

Q25. Solution:

• ∫₀^{∞} e^{−ax} dx = 1/a for a>0. With a=1.25 ⇒ 0.80.