More practice questions

5. Q5. Evaluate ∫₀¹ x(1 − x)ⁿ dx for real n > −1 (NCERT Exercise 7.10, Q7, page 280).

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   By P_₄, x → 1 − x: the integral is ∫₀¹ (1 − x)xⁿ dx = 1/(n + 1) − 1/(n + 2) = 1/((n + 1)(n + 2)).

6. Q6. Evaluate ∫₀² x√(2 − x) dx (NCERT Exercise 7.10, Q9, page 280).

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   By P_₄, x → 2 − x: the integral is ∫₀² (2 − x)√x dx = [(4/3)x^3/2 − (2/5)x^5/2] from 0 to 2 = 8√2/3 − 8√2/5 = 16√2/15.

7. Q7. Evaluate ∫₀^π/2 (2 log sin x − log sin 2x) dx (NCERT Exercise 7.10, Q10, page 280).

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   All the endpoint logarithmic integrals converge. On the open interval, log sin 2x = log 2 + log sin x + log cos x, so the integrand is log sin x − log cos x − log 2. By P_₄ the first two integrals are equal, so the value is −(π/2) log 2, which NCERT writes as (π/2) log (1/2).

8. Q8. Evaluate ∫_−π/2^π/2 sin⁷ x dx and ∫₀^2π cos⁵ x dx (NCERT Exercise 7.10, Q13 and Q14, page 280).

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   sin⁷ x is odd, so the first is 0 by P_₇. For the second, cos⁵ (2π − x) = cos⁵ x, so by P_₆ it is 2∫₀^π cos⁵ x dx; and cos⁵ (π − x) = −cos⁵ x, so by P_₆ again that is 0.

9. Q9. Evaluate ∫₀^π/2 (sin x − cos x)/(1 + sin x cos x) dx (NCERT Exercise 7.10, Q15, page 280).

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   By P_₄, sin and cos swap, so the top changes sign and the bottom does not: I = −I, and I = 0.

10. Q10. Evaluate ∫₀^a √x/(√x + √(a − x)) dx with a > 0 (NCERT Exercise 7.10, Q17, page 280).

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    By P_₄, x → a − x gives the partner √(a − x)/(√(a − x) + √x). Adding, 2I = ∫₀^a 1 dx = a, so I = a/2.

11. Q11. Evaluate ∫₀⁴ |x − 1| dx (NCERT Exercise 7.10, Q18, page 280).

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    Split at 1: ∫₀¹ (1 − x) dx + ∫₁⁴ (x − 1) dx = ½ + 9/2 = 5.

12. Q12. Show that ∫₀^a f(x)g(x) dx = 2∫₀^a f(x) dx if f(x) = f(a − x) and g(x) + g(a − x) = 4 (NCERT Exercise 7.10, Q19, page 280).

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    Let I be the left side. By P_₄, I = ∫₀^a f(a − x)g(a − x) dx = ∫₀^a f(x)(4 − g(x)) dx = 4∫₀^a f(x) dx − I. So 2I = 4∫₀^a f(x) dx, which is the result.

13. Q13. Beyond the NCERT text (JEE): evaluate ∫₋₁¹ dx/(1 + e^x).

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    Equation (1) in NCERT’s proof of P_₇ (page 275) holds for any f: the integral over [−a, a] is ∫₀^a (f(x) + f(−x)) dx. Here f(−x) = e^x/(1 + e^x), so f(x) + f(−x) = 1, and the value is ∫₀¹ 1 dx = 1.

14. Q14. Beyond the NCERT text (JEE): evaluate ∫₀^4π |sin x| dx.

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    |sin x| repeats every π, and over each of the four intervals of length π its integral equals ∫₀^π sin x dx = 2. So by P_₂ the value is 4 × 2 = 8.