Evaluate the following integral as the limit of sum: ∫ 0 2 e x d x .

Let I = ∫ 0 2 e x d x f ( x ) = e x is continuous in [ 0 , 2 ] Using definite integral as a limit of sum we get, I = ∫ 0 2 e x d x = lim n → ∞ 2 n ∑ r = 0 n - 1 f 2 r n I = lim n → ∞ 2 n ∑ r = 0 n - 1 e 2 r n Using sum to n terms of Geometric Progression we get, I = lim n → ∞ 2 n 1 - e 2 n n 1 - e 2 n I = lim n → ∞ 2 n 1 - e 2 1 - e 2 n Let 2 n = z Then as, n → ∞ ⇒ z → 0 So we get, I = ( 1 - e 2 ) lim z → 0 z 1 - e z Using the standard Limit, lim x → 0 e x - 1 x = 1 we get, I = e 2 - 1

Evaluate the following integral as the limit of sum: ∫ 0 2 e x d x .

Let I = ∫ 0 2 e x d x f ( x ) = e x is continuous in [ 0 , 2 ] Using definite integral as a limit of sum we get, I = ∫ 0 2 e x d x = lim n → ∞ 2 n ∑ r = 0 n - 1 f 2 r n I = lim n → ∞ 2 n ∑ r = 0 n - 1 e 2 r n Using sum to n terms of Geometric Progression we get, I = lim n → ∞ 2 n 1 - e 2 n n 1 - e 2 n I = lim n → ∞ 2 n 1 - e 2 1 - e 2 n Let 2 n = z Then as, n → ∞ ⇒ z → 0 So we get, I = ( 1 - e 2 ) lim z → 0 z 1 - e z Using the standard Limit, lim x → 0 e x - 1 x = 1 we get, I = e 2 - 1

Evaluate the following integral as the limit of sum: ∫ a b cos x d x

Let I = ∫ a b ( cos x ) d x f ( x ) = cos x is continuous in [ a , b ] Now by the formula of definite integral as a limit of a sum, I = ∫ a b ( cos x ) d x = lim n → ∞ h ∑ r = 0 n - 1 f ( a + r h ) , where h = b - a n ⇒ I = lim n → ∞ b - a n ∑ r = 0 n - 1 cos ( a + r h ) . . . . 1 Now consider the summation, S = cos ( a ) + cos ( a + h ) + cos ( a + 2 h ) + cos ( a + 3 h ) + … … … … + cos ( a + ( n - 1 ) h ) Multiply both sides with 2 sin h 2 and use, 2 cos A sin B = sin ( A + B ) - sin ( A - B ) So we get, 2 S sin h 2 = sin a + h 2 - sin a - h 2 + sin a + 3 h 2 - sin a + h 2 + sin a + 5 h 2 - sin a + 3 h 2 + ⋯ sin a + ( 2 n - 1 ) h 2 - sin a + ( 2 n - 3 ) h 2 cancelling some terms we get, 2 S sin h 2 = sin a + ( 2 n - 1 ) h 2 - sin a - h 2 Now using, S i n C - sin D = 2 cos C + D 2 sin C - D 2 we get, S = sin n h 2 cos a + ( n - 1 ) h 2 sin h 2 Putting h = b - a n I = lim n → ∞ b - a n sin n ( b - a ) 2 n cos a + ( n - 1 ) ( b - a ) 2 n sin b - a 2 n b - a 2 n · b - a 2 n As we know lim h → 0 sin h h = 1 I = lim n → ∞ 2 sin ( b - a ) 2 cos a + 1 2 - 1 2 n ( b - a ) I = 2 sin b - a 2 cos b + a 2 Hence, I = sin ( b ) - sin ( a ) .