Find the value of ∫ 0 π 3 sin x d x using first fundamental theorem of integral calculus.

Given, ∫ 0 π 3 sin x d x If f x is continuous on interval a , b and F x is any function that satisfies F ' x = f x , then ∫ a b f x d x = F b - F a . Let f x = sin x We know that, d d x - cos x = sin x ⇒ F x = - cos x ∴ ∫ 0 π 3 sin x d x = - cos π 3 - - cos 0 = - 1 2 - - 1 ∴ ∫ 0 π 3 sin x d x = 1 2

Find the value of ∫ 0 π 4 cos x d x using first fundamental theorem of integral calculus.

Given, ∫ 0 π 4 cos x d x If f x is continuous on interval a , b and F x is any function that satisfies F ' x = f x , then ∫ a b f x d x = F b - F a . Let f x = cos x We know that, d d x sin x = cos x ⇒ F x = sin x ∴ ∫ 0 π 4 cos x d x = sin π 4 - sin 0 = 1 2 - 0 ∴ ∫ 0 π 4 cos x d x = 1 2

Find the value of ∫ 0 π 2 sin x d x using first fundamental theorem of integral calculus.

Given, ∫ 0 π 2 sin x d x If f x is continuous on interval a , b and F x is any function that satisfies F ' x = f x , then ∫ a b f x d x = F b - F a . Let f x = sin x We know that, d d x - cos x = sin x ⇒ F x = - cos x ∴ ∫ 0 π 2 sin x d x = - cos π 2 - - cos 0 = 0 - - 1 ∴ ∫ 0 π 2 sin x d x = 1

Evaluate the integral: ∫ 0 2 x 2 - x d x .

Let I = ∫ 0 2 x 2 - x d x Using the property of definite integrals , ∫ a b f x d x = ∫ a b f a + b - x d x Now apply the property, we get I = ∫ 0 2 ( 2 - x ) x d x ⇒ I = ∫ 0 2 2 x d x - ∫ 0 2 x 3 2 d x ⇒ I = ∫ 0 2 2 x 1 2 d x - ∫ 0 2 x 3 2 d x ⇒ I = 2 × 2 3 x 3 2 - 2 5 x 5 2 0 2 Now apply upper and lower limits respectively, so we get: ⇒ I = 2 × 2 3 2 3 2 - 2 5 2 5 2 - 0 ⇒ I = 2 2 4 3 - 4 5 ⇒ I = 16 15 2 Hence, ∫ 0 2 x 2 - x d x = 16 15 2 .

Evaluate: ∫ - 1 3 2 x sin π x d x .

Let I = ∫ - 1 3 2 x sin π x d x We know that, | x · sin π x | = x · sin π x , - 1 ≤ x ≤ 1 and | x · sin π x | = - x sin π x , 1 < x ≤ 3 2 ∴ I = ∫ - 1 3 2 | x sin π x | d x = ∫ - 1 1 | x sin π x | d x + ∫ 1 3 2 | x sin π x | d x ⇒ I = ∫ - 1 1 x sin π x d x - ∫ 1 3 2 x sin π x d x ⇒ I = x ∫ - 1 1 sin π x d x - ∫ - 1 1 d d x x ∫ - 1 1 sin π x d x d x - x ∫ 1 3 2 sin π x d x - ∫ 1 3 2 d d x x ∫ 1 3 2 sin π x d x d x (Using Integration By Parts) ⇒ I = - x · cos π x π + sin π x π 2 - 1 1 - - x · cos π x π + sin π x π 2 1 3 2 ⇒ I = - 1 · cos π π + sin π π 2 - 1 · cos π π - sin - π π 2 - - 3 2 · cos π 3 2 π + sin π 3 2 π 2 + 1 · cos π π - 0 ⇒ I = 1 π + ( - 1 ) ( - 1 ) π - - 1 π 2 - 1 π ⇒ I = 1 π + 1 π + 1 π 2 + 1 π ⇒ I = 3 π + 1 π 2 Hence, ∫ - 1 3 2 | x sin π x | d x = 3 π + 1 π 2 .

Evaluate the integral: ∫ 0 a x ( a - x ) n d x .

Let I = ∫ 0 a x ( a - x ) n d x ⇒ I = ∫ 0 a ( a - x ) ( x ) n d x ∵ ∫ a b f ( x ) d x = ∫ a b f ( a + b - x ) d x ⇒ I = ∫ 0 a a x n d x - ∫ 0 a x n + 1 d x ⇒ I = a x n + 1 n + 1 - x n + 2 n + 2 0 a ⇒ I = a · a n + 1 n + 1 - a n + 2 n + 2 - 0 ⇒ I = a n + 2 n + 1 - a n + 2 n + 2 ⇒ I = a n + 2 ( n + 1 ) ( n + 2 ) Hence, ∫ 0 a x ( a - x ) n d x = a n + 2 ( n + 1 ) ( n + 2 ) .

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Calculus>Chapter 7 - Definite Integrals>Exercise

Embibe Experts Solutions for Exercise 1: Exercise

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Exercise 1: Exercise

Attempt the practice questions from Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.