Take timed practice tests on Definite Integration and Area for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.
Six 20-question timed sessions plus one 60-question module test. Each question is original and calibrated from the uploaded material pattern without copying PDF wording.
Ready
1. ∫₀¹ x² dx equals:
Explanation: ∫₀¹ x² dx = [x³/3]₀¹ = 1/3 − 0 = 1/3.
2. ∫ₐᵇ f(x) dx = −∫ᵦₐ f(x) dx. This property states:
Explanation: Swapping the limits of integration reverses the sign: ∫ₐᵇ f(x) dx = −∫ᵦₐ f(x) dx. This is a fundamental property of definite integrals.
3. The area under y = f(x) from x = a to x = b (f(x) ≥ 0) is:
Explanation: When f(x) ≥ 0 on [a,b], the area under the curve = ∫ₐᵇ f(x) dx. When f(x) can be negative, area = ∫ₐᵇ |f(x)| dx.
4. If f(x) is an odd function, then ∫₋ₐᵃ f(x) dx equals:
Explanation: For odd function f(−x) = −f(x): ∫₋ₐᵃ f(x) dx = ∫₋ₐ⁰ f(x) dx + ∫₀ᵃ f(x) dx = −∫₀ᵃ f(x) dx + ∫₀ᵃ f(x) dx = 0.
5. ∫₀^(π/2) sin x/(sin x + cos x) dx equals:
Explanation: Using King's property: I = ∫₀^(π/2) cos x/(cos x + sin x) dx. Adding: 2I = ∫₀^(π/2) 1 dx = π/2. So I = π/4.
6. ∫₀¹ eˣ dx equals:
Explanation: ∫₀¹ eˣ dx = [eˣ]₀¹ = e¹ − e⁰ = e − 1.
7. The area between y = x² and y = x from x = 0 to x = 1 is:
Explanation: On [0,1]: x ≥ x². Area = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.
8. If f(x) has period T, then ∫₀^(nT) f(x) dx equals:
Explanation: For a function with period T: ∫₀^(nT) f(x) dx = n × ∫₀ᵀ f(x) dx. Each complete period contributes the same integral.
9. ∫₀^π x sin x dx equals:
Explanation: By parts: ∫₀^π x sin x dx = [−x cos x]₀^π + ∫₀^π cos x dx = (−π cos π + 0) + [sin x]₀^π = π + 0 = π.
10. d/dx ∫ₐ^(g(x)) f(t) dt equals:
Explanation: Leibniz differentiation under the integral sign: d/dx ∫ₐ^(g(x)) f(t) dt = f(g(x)) × g'(x) (by the chain rule applied to the Fundamental Theorem of Calculus).
11. ∫₋₁¹ x|x| dx equals:
Explanation: f(x) = x|x|. For x ≥ 0: x². For x
12. The area enclosed by y = |x| and y = 1 is:
Explanation: y = |x| and y = 1 intersect at x = ±1. Area = ∫₋₁¹ (1 − |x|) dx = 2∫₀¹(1−x) dx = 2[x−x²/2]₀¹ = 2(1/2) = 1 sq unit.
13. ∫₀² (2x + 1) dx equals:
Explanation: ∫₀²(2x+1) dx = [x²+x]₀² = (4+2) − 0 = 6.
14. The area of the region bounded by y = x², x-axis, x = 0, and x = 3 is:
Explanation: Area = ∫₀³ x² dx = [x³/3]₀³ = 27/3 = 9 sq units.
15. ∫₀^(π/2) sin²x dx equals:
Explanation: sin²x = (1−cos 2x)/2. ∫₀^(π/2) (1−cos 2x)/2 dx = [x/2 − sin 2x/4]₀^(π/2) = (π/4 − 0) − 0 = π/4.