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.
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1. ā«āĀ¹ xĀ² dx equals:
Explanation: ā«āĀ¹ xĀ² dx = [xĀ³/3]āĀ¹ = 1/3 ā 0 = 1/3.
2. ā«āįµ f(x) dx = āā«įµ¦ā f(x) dx. This property states:
Explanation: Swapping the limits of integration reverses the sign: ā«āįµ f(x) dx = āā«įµ¦ā 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) ā„ 0) is:
Explanation: When f(x) ā„ 0 on [a,b], the area under the curve = ā«āįµ f(x) dx. When f(x) can be negative, area = ā«āįµ |f(x)| dx.
4. If f(x) is an odd function, then ā«āāįµ f(x) dx equals:
Explanation: For odd function f(āx) = āf(x): ā«āāįµ f(x) dx = ā«āāā° f(x) dx + ā«āįµ f(x) dx = āā«āįµ f(x) dx + ā«āįµ f(x) dx = 0.
5. ā«ā^(Ļ/2) sin x/(sin x + cos x) dx equals:
Explanation: Using King's property: I = ā«ā^(Ļ/2) cos x/(cos x + sin x) dx. Adding: 2I = ā«ā^(Ļ/2) 1 dx = Ļ/2. So I = Ļ/4.
6. ā«āĀ¹ eĖ£ dx equals:
Explanation: ā«āĀ¹ eĖ£ dx = [eĖ£]āĀ¹ = eĀ¹ ā eā° = e ā 1.
7. The area between y = xĀ² and y = x from x = 0 to x = 1 is:
Explanation: On [0,1]: x ā„ xĀ². Area = ā«āĀ¹ (x ā xĀ²) dx = [xĀ²/2 ā xĀ³/3]āĀ¹ = 1/2 ā 1/3 = 1/6.
8. If f(x) has period T, then ā«ā^(nT) f(x) dx equals:
Explanation: For a function with period T: ā«ā^(nT) f(x) dx = n Ć ā«āįµ f(x) dx. Each complete period contributes the same integral.
9. ā«ā^Ļ x sin x dx equals:
Explanation: By parts: ā«ā^Ļ x sin x dx = [āx cos x]ā^Ļ + ā«ā^Ļ cos x dx = (āĻ cos Ļ + 0) + [sin x]ā^Ļ = Ļ + 0 = Ļ.
10. d/dx ā«ā^(g(x)) f(t) dt equals:
Explanation: Leibniz differentiation under the integral sign: d/dx ā«ā^(g(x)) f(t) dt = f(g(x)) Ć g'(x) (by the chain rule applied to the Fundamental Theorem of Calculus).
11. ā«āāĀ¹ x|x| dx equals:
Explanation: f(x) = x|x|. For x ā„ 0: xĀ². For x
12. The area enclosed by y = |x| and y = 1 is:
Explanation: y = |x| and y = 1 intersect at x = Ā±1. Area = ā«āāĀ¹ (1 ā |x|) dx = 2ā«āĀ¹(1āx) dx = 2[xāxĀ²/2]āĀ¹ = 2(1/2) = 1 sq unit.
13. ā«āĀ² (2x + 1) dx equals:
Explanation: ā«āĀ²(2x+1) dx = [xĀ²+x]āĀ² = (4+2) ā 0 = 6.
14. The area of the region bounded by y = xĀ², x-axis, x = 0, and x = 3 is:
Explanation: Area = ā«āĀ³ xĀ² dx = [xĀ³/3]āĀ³ = 27/3 = 9 sq units.
15. ā«ā^(Ļ/2) sinĀ²x dx equals:
Explanation: sinĀ²x = (1ācos 2x)/2. ā«ā^(Ļ/2) (1ācos 2x)/2 dx = [x/2 ā sin 2x/4]ā^(Ļ/2) = (Ļ/4 ā 0) ā 0 = Ļ/4.