Evaluate ∫ 0 2 x 3 d x as limit of sum.

Let I = ∫ 0 2 x 3 d x By using first principle, ∫ a b f ( x ) d x = lim h → 0 h ∑ r = 1 n f ( a + r h ) where n h = b - a & n → ∞ Here a = 0 , b = 2 ∴ n h = b - a = 2 - 0 = 2 And f ( x ) = x 3 ⇒ f ( a + r h ) = f ( 0 + r h ) = ( r h ) 3 ∴ I = ∫ 0 2 x 3 d x = lim h → 0 h ∑ r = 1 n r 3 h 3 = lim h → 0 h 4 n ( n + 1 ) 2 2 = lim h → 0 n h ( n h + h ) 2 2 { ∵ n h = 2 } = lim h → 0 2 ( 2 + h ) 2 2 ∴ I = 4

Evaluate ∫ 0 2 e x d x as limit of sum.

Let I = ∫ 0 2 e x d x By using first principle, ∫ a b f ( x ) d x = lim h → 0 h ∑ r = 1 n f ( a + r h ) where n h = b - a & n → ∞ Here a = 0 , b = 2 ∴ n h = b - a = 2 - 0 = 2 And f ( x ) = e x ⇒ f ( a + r h ) = f ( 0 + r h ) = e r h I = ∫ 0 2 e x d x I = lim h → 0 h ∑ r = 1 n e r h I = lim h → 0 h e h + e 2 h + . . . + e n h I = lim h → 0 h e h e h n - 1 e h - 1 I = lim h → 0 h e h e 2 - 1 h e h - 1 h { ∵ n h = 2 } I = lim h → 0 e h ( e 2 - 1 ) e h - 1 h I = e 2 - 1 1 I = e 2 - 1

Evaluate ∫ 1 3 e - x d x as limit of sum.

Let I = ∫ 1 3 e - x d x By using first principle, ∫ a b f x d x = lim h → 0 h ∑ r = 1 n f a + r h where n h = b - a & n → ∞ Here a = 1 , b = 3 ∴ n h = b - a = 3 - 1 = 2 f ( x ) = e - x ⇒ f ( a + r h ) = f ( 1 + r h ) = e - ( 1 + r h ) = e - 1 . e - r h I = ∫ 1 3 e - x d x ⇒ I = lim h → 0 h ∑ r = 1 n e - 1 e - r h I = lim h → 0 h e - 1 e - h [ ( e - h ) n - 1 ] e - h - 1 = lim h → 0 e - 1 e - h e - 2 - 1 - e - h - 1 - h = e - 1 ( e - 2 - 1 ) - 1 ∵ n h = 2 ∴ I = e - 1 - e - 3

Evaluate ∫ a b e x dx as limit of sum.

Let I = ∫ a b e x d x By using first principle, ∫ a b f x d x = lim h → 0 h ∑ r = 1 n f a + r h where n h = b - a & n → ∞ Here a = a , b = b ∴ n h = b - a And f ( x ) = e x ⇒ f ( a + r h ) = e a + r h = e a . e r h I = ∫ a b e x d x = lim h → 0 h ∑ r = 1 n e a . e r h I = lim h → 0 h . e a e h + e 2 h + e 3 h + . . . + e n h I = lim h → 0 h . e a e h e h n - 1 e h - 1 I = lim h → 0 e a . e h [ e b - a - 1 ] e h - 1 h ∵ n h = b - a I = e a ( e b - a - 1 ) I = e b - e a

Evaluate ∫ 0 1 e 2 - 3 x d x as limit of sum.

Assume: I = ∫ 0 1 e 2 - 3 x d x By using first principle, ∫ a b f ( x ) d x = lim h → 0 h ∑ r = 1 n f ( a + r h ) Where, n h = b - a & n → ∞ Here, a = 0 , b = 1 ∴ n h = b - a = 1 And f ( x ) = e 2 - 3 x = e 2 · e - 3 x ⇒ f ( a + r h ) = f ( 0 + r h ) = e 2 · e - 3 ( r h ) I = ∫ 0 1 e 2 - 3 x d x = lim h → 0 h ∑ r = 1 n e 2 · e - 3 r h I = lim h → 0 h · e 2 e - 3 h + e - 6 h + e - 9 h + . . . + e - 3 n h I = lim h → 0 h · e 2 e - 3 h e - 3 h n - 1 e - 3 h - 1 I = lim h → 0 e 2 · e - 3 h [ e - 3 - 1 ] ( - 3 ) e - 3 h - 1 - 3 h I = e 2 [ e - 3 - 1 ] - 3 { ∵ n h = 1 } I = e 2 - e - 1 3

Evaluate ∫ 2 4 2 x d x as limit of sum.

Assume: I = ∫ 2 4 2 x d x By using first principle, ∫ a b f x d x = lim h → 0 h ∑ r = 1 n f a + r h Where, n h = b - a & n → ∞ Here, a = 2 , b = 4 ∴ n h = b - a = 4 - 2 = 2 And f ( x ) = 2 x ⇒ f ( a + r h ) = f ( 2 + r h ) = 2 2 + r h = 2 2 · 2 r h = 4 · 2 r h I = ∫ 2 4 2 x d x = lim h → 0 h ∑ r = 1 n 4 · 2 r h I = lim h → 0 4 h 2 h + 2 2 h + . . . + 2 n h I = lim h → 0 4 h 2 h 2 h n - 1 2 h - 1 ∵ n h = 2 I = lim h → 0 4 · 2 h 2 2 - 1 2 h - 1 h I = 4 × 3 log 2 I = 12 log 2

Evaluate ∫ a b e m x d x as limit of sum.

Assume: I = ∫ a b e m x d x By using first principle, ∫ a b f ( x ) d x = lim h → 0 h ∑ r = 1 n f ( a + r h ) where, n h = b - a & n → ∞ Here, a = a , b = b ∴ n h = b - a And f ( x ) = e m x ⇒ f ( a + r h ) = e m ( a + r h ) = e m a · e m r h I = ∫ a b e m x d x = lim h → 0 h ∑ r = 1 n e m a · e m r h I = lim h → 0 h · e m a e m h + e 2 m h + . . . + e n m h I = lim h → 0 h · e m a e m h e m h n - 1 e m h - 1 I = lim h → 0 e m a · e m h [ e m ( b - a ) - 1 ] m e m h - 1 m h { ∵ n h = b - a } I = e m a · [ e m b - m a - 1 ] m I = e m b - e m a m

Evaluate ∫ 0 π 2 cos x d x as limit of sum.

Assume: I = ∫ 0 π 2 cos x d x By using first principle, ∫ a b f ( x ) d x = lim h → 0 h ∑ r = 1 n f ( a + r h ) where, n h = b - a & n → ∞ Here, a = 0 , b = π 2 ∴ n h = π 2 - 0 = π 2 And f ( x ) = cos x ⇒ f ( a + r h ) = f ( 0 + r h ) = cos ( r h ) I = ∫ 0 π 2 cos d x = lim h → 0 h ∑ r = 1 n cos ( r h ) I = lim h → 0 h cos h + cos 2 h + . . . + cos n h I = lim h → 0 h sin n · h 2 cos h + ( n - 1 ) · h 2 sin h 2 I = lim h → 0 sin π 2 2 cos h 2 + π 2 2 1 2 sin h 2 h 2 ∵ n h = π 2 I = 2 1 2 × 1 2 I = 1

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Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced>Chapter 3 - Properties of Definite Integrals>EXERCISE

K. C. Sinha Solutions for Chapter: Properties of Definite Integrals, Exercise 2: EXERCISE

Author: K. C. Sinha

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha Solutions for Chapter: Properties of Definite Integrals, Exercise 2: EXERCISE

Attempt the free practice questions on Chapter 3: Properties of Definite Integrals, Exercise 2: EXERCISE with hints and solutions to strengthen your understanding. Eduwiser's INTEGRAL CALCULUS For JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.