PYQs Learn Notes Mocks Board Questions Maharashtra HSC Class 12 Mathematics Definite Integration 15 Maharashtra HSC Class 12 Mathematics previous year questions (PYQs) from 9 papers, 2016 to 2026, with answers and worked solutions.
About this chapter Difficulty: 4 easy · 6 moderate · 5 hard. Most-asked subtopics: Properties of Definite Integrals (9), Fundamental Theorem of Integral Calculus (3), Methods of Evaluation of Definite Integrals (3). Updated 19 September 2026. Try each question before you open its answer. Sign in, and the ones you get wrong come back in Fix your mistakes until you get them right.
#1 · Properties of Definite Integrals
Evaluate : ∫ − 1 1 ∣ x ∣ d x \int_{-1}^{1} |x| \, dx ∫ − 1 1 ∣ x ∣ d x #2 · Properties of Definite Integrals
Prove that,
∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x \int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x , if f ( x ) f(x) f ( x ) is even
= 0 = 0 = 0 , if f ( x ) f(x) f ( x ) is odd #3 · Properties of Definite Integrals
Prove that :
∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x \int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) \, dx ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x if f f f is an even function
= 0 = 0 = 0 , if f f f is an odd function
Hence find the value of ∫ − 1 1 tan − 1 x d x \int_{-1}^{1} \tan^{-1} x \, dx ∫ − 1 1 tan − 1 x d x . #4 · Fundamental Theorem of Integral Calculus
Evaluate : ∫ 0 − 1 e − x d x \int_{0}^{-1} e^{-x} \, dx ∫ 0 − 1 e − x d x #5 · Methods of Evaluation of Definite Integrals
Evaluate : ∫ 0 1 2 d x ( 1 − 2 x 2 ) ⋅ 1 − x 2 \int_{0}^{\frac{1}{2}} \frac{dx}{(1 - 2x^{2}) \cdot \sqrt{1 - x^{2}}} ∫ 0 2 1 ( 1 − 2 x 2 ) ⋅ 1 − x 2 d x Done a few? Here’s your next move
#6 · Methods of Evaluation of Definite Integrals
∫ 1 2 1 x 2 ⋅ e 1 x d x = \int_{1}^{2} \frac{1}{x^{2}} \cdot e^{\frac{1}{x}} dx = ∫ 1 2 x 2 1 ⋅ e x 1 d x = _____. A B C e ( e − 1 ) \sqrt{e}(\sqrt{e} - 1) e ( e − 1 ) D e − 1 e \frac{\sqrt{e} - 1}{e} e e − 1 #7 · Properties of Definite Integrals
Prove that : ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) ⋅ d x \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) \cdot dx ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) ⋅ d x
Hence, find ∫ π 6 π 3 sin 2 x ⋅ d x \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sin^{2} x \cdot dx ∫ 6 π 3 π sin 2 x ⋅ d x Maharashtra HSC Class 12 · 2025 · Q. 12 Moderate Add #8 · Fundamental Theorem of Integral Calculus
Evaluate: ∫ − π / 4 π / 4 1 1 − s i n x d x \int_{- \pi/4}^{\pi/4}\frac{1}{1 - sinx}dx ∫ − π /4 π /4 1 − s in x 1 d x . Maharashtra HSC Class 12 · 2025 · Q. 33 Moderate Add #9 · Properties of Definite Integrals
Prove that: ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int_{a}^{b}f(x)dx = \int_{a}^{b}f(a + b - x)dx ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x . Hence evaluate: ∫ 0 3 x x + 3 − x d x \int_{0}^{3}\frac{\sqrt{x}}{\sqrt{x} + \sqrt{3 - x}}dx ∫ 0 3 x + 3 − x x d x . Maharashtra HSC Class 12 · 2024 · Q. 1. vi. Easy Add #10 · Properties of Definite Integrals
If ∫ − π / 4 π / 4 x 3 sin 4 x d x = k \int_{- \pi/4}^{\pi/4}x^{3}\sin^{4}xdx = k ∫ − π /4 π /4 x 3 sin 4 x d x = k then k = __________. Maharashtra HSC Class 12 · 2024 · Q. 11 Easy Add #11 · Fundamental Theorem of Integral Calculus
Evaluate: ∫ 0 π 2 cos 2 x d x \int_{0}^{\frac{\pi}{2}}\cos^{2}xdx ∫ 0 2 π cos 2 x d x . Maharashtra HSC Class 12 · 2024 · Q. 34 Moderate Add #12 · Properties of Definite Integrals
Prove that: ∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x \int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)dx + \int_{0}^{a}f(2a - x)dx ∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x . Hence show that: ∫ 0 π sin x d x = 2 ∫ 0 π / 2 sin x d x \int_{0}^{\pi}\sin xdx = 2\int_{0}^{\pi/2}\sin xdx ∫ 0 π sin x d x = 2 ∫ 0 π /2 sin x d x . Maharashtra HSC Class 12 · 2022 · Q. 34 Hard Add #13 · Properties of Definite Integrals
Show that: ∫ 0 π log ( 1 + t a n x ) d x = π 8 l o g 2 \int_{0}^{\pi}\log(1 + tanx)dx = \frac{\pi}{8}log2 ∫ 0 π log ( 1 + t an x ) d x = 8 π l o g 2 . Maharashtra HSC Class 12 · 2018 · Q. 4. (A) ii. Moderate Add #14 · Methods of Evaluation of Definite Integrals
If ∫ 0 k 1 2 + 8 x 2 d x = π 16 \int_{0}^{k}\frac{1}{2 + 8x^{2}}dx = \ \frac{\pi}{16} ∫ 0 k 2 + 8 x 2 1 d x = 16 π , then the value of k k k is ______. Maharashtra HSC Class 12 · 2016 · Q. 6. (A) i. Easy Add #15 · Properties of Definite Integrals
Show that: ∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x \int_{0}^{2a}f(x)dx = \int_{0}^{a}f(x)dx + \int_{0}^{a}f(2a - x)dx ∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x . Studying this with friends? Send them this chapter's questions. It's free for them too.
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