Evaluate ∫ 1 3 2 x + 1 d x as limit of sum.

Let I = ∫ 1 3 2 x + 1 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 f ( x ) = 2 x + 1 and a = 1 , b = 3 ∴ n h = b - a = 3 - 1 = 2 & f ( a + r h ) = f ( 1 + r h ) = 2 ( 1 + r h ) + 1 = 2 r h + 3 ∴ I = ∫ 1 3 2 x + 1 d x = lim h → 0 h ∑ r = 1 n 2 r h + 3 = lim h → 0 2 h 2 ∑ r = 1 n r + 3 h ∑ r = 1 n 1 = lim h → 0 2 h 2 n ( n + 1 ) 2 + 3 n h = lim h → 0 n h ( n h + h ) + 3 n h = lim h → 0 2 ( 2 + h ) + 3 × 2 ∵ n h = 2 = 4 + 6 ∴ I = 10

Evaluate ∫ 1 3 2 x + 1 d x as limit of sum.

Let I = ∫ 1 3 2 x + 1 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 f ( x ) = 2 x + 1 and a = 1 , b = 3 ∴ n h = b - a = 3 - 1 = 2 & f ( a + r h ) = f ( 1 + r h ) = 2 ( 1 + r h ) + 1 = 2 r h + 3 ∴ I = ∫ 1 3 2 x + 1 d x = lim h → 0 h ∑ r = 1 n 2 r h + 3 = lim h → 0 2 h 2 ∑ r = 1 n r + 3 h ∑ r = 1 n 1 = lim h → 0 2 h 2 n ( n + 1 ) 2 + 3 n h = lim h → 0 n h ( n h + h ) + 3 n h = lim h → 0 2 ( 2 + h ) + 3 × 2 ∵ n h = 2 = 4 + 6 ∴ I = 10

Evaluate ∫ a b x d x as limit of sum.

Let I = ∫ a b 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 f ( x ) = x and a = a , b = b ∴ n h = b - a and f ( a + r h ) = ( a + r h ) ∴ I = ∫ a b f x d x = lim h → 0 h ∑ r = 1 n a + r h = lim h → 0 h 2 ∑ r = 1 n r + a h ∑ r = 1 n 1 = lim h → 0 h 2 n ( n + 1 ) 2 + a h n = lim h → 0 n h ( n h + h ) 2 + a n h = lim h → 0 ( b - a ) ( b - a + h ) 2 + a b - a { ∵ n h = b - a } = ( b - a ) 2 2 + a b - a = b - a b - a 2 + a = 1 2 b - a b + a ∴ I = 1 2 ( b 2 - a 2 )

Evaluate ∫ 1 2 x 2 d x as limit of sum.

Let I = ∫ 1 2 x 2 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 f ( x ) = x 2 and a = 1 , b = 2 ∴ n h = b - a = 2 - 1 = 1 Now f ( a + r h ) = f ( 1 + r h ) = ( 1 + r h ) 2 = r 2 h 2 + 2 r h + 1 ∴ I = ∫ 1 2 x 2 d x = lim h → 0 h ∑ r = 1 n r 2 h 2 + 2 r h + 1 = lim h → 0 h 3 ∑ r = 1 n r 2 + 2 h 2 ∑ r = 1 n r + h ∑ r = 1 n 1 = lim h → 0 h 3 n ( n + 1 ) ( 2 n + 1 ) 6 + 2 · h 2 n ( n + 1 ) 2 + h n = lim h → 0 n h ( n h + h ) ( 2 n h + h ) 6 + n h n h + h + n h = lim h → 0 1 ( 1 + h ) ( 2 + h ) 6 + 1 ( 1 + h ) + 1 { ∵ n h = 1 } = 2 6 + 1 + 1 ∴ I = 7 3

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

Let I = ∫ 2 3 x 2 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 f ( x ) = x 2 and a = 2 , b = 3 ∴ n h = b - a = 3 - 2 = 1 Now f ( a + r h ) = f ( 2 + r h ) = ( 2 + r h ) 2 = r 2 h 2 + 4 r h + 4 ∴ I = ∫ 2 3 x 2 d x = lim h → 0 h ∑ r = 1 n ( r 2 h 2 + 4 r h + 4 ) = lim h → 0 h 3 ∑ r = 1 n r 2 + 4 h 2 ∑ r = 1 n r + 4 h ∑ r = 1 n 1 = lim h → 0 h 3 n ( n + 1 ) ( 2 n + 1 ) 6 + 4 h 2 n ( n + 1 ) 2 + 4 h n = lim h → 0 n h ( n h + h ) ( 2 n h + h ) 6 + 2 n h n h + h + 4 n h = lim h → 0 1 ( 1 + h ) ( 2 + h ) 6 + 2 ( 1 + h ) + 4 { ∵ n h = 1 } = 2 6 + 2 + 4 ∴ I = 19 3

Evaluate ∫ 0 2 ( x 2 + 1 ) d x as limit of sum.

Let I = ∫ 0 2 ( x 2 + 1 ) 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 & f ( x ) = x 2 + 1 ⇒ f ( a + r h ) = f ( 0 + r h ) = ( r h ) 2 + 1 ∴ I = ∫ 0 2 ( x 2 + 1 ) d x = lim h → 0 h ∑ r = 1 n ( r 2 h 2 + 1 ) = lim h → 0 h 3 ∑ r = 1 n r 2 + h ∑ r = 1 n 1 = lim h → 0 h 3 n ( n + 1 ) ( 2 n + 1 ) 6 + h n = lim h → 0 n h ( n h + h ) ( 2 n h + h ) 6 + n h = lim h → 0 2 ( 2 + h ) ( 2 × 2 + h ) 6 + 2 { ∵ n h = 2 } = 16 6 + 2 ∴ I = 14 3

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

Let I = ∫ 0 2 ( x 2 + 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 And f ( x ) = x 2 + 3 ⇒ f ( a + r h ) = f ( 0 + r h ) = ( r h ) 2 + 3 ∴ I = ∫ 0 2 ( x 2 + 3 ) d x = lim h → 0 h ∑ r = 1 n ( r 2 h 2 ) + 3 = lim h → 0 h 3 ∑ r = 1 n r 2 + 3 h ∑ r = 1 n 1 = lim h → 0 h 3 n ( n + 1 ) ( 2 n + 1 ) 6 + 3 h n = lim h → 0 n h ( n h + h ) ( 2 n h + h ) 6 + 3 n h = lim h → 0 2 ( 2 + h ( 2 × 2 + h ) 6 + 3 × 2 { ∵ n h = 2 } = 16 6 + 6 ∴ I = 26 3

Evaluate ∫ 1 3 ( 2 x 2 + 5 ) d x as limit of sum.

Let I = ∫ 1 3 ( 2 x 2 + 5 ) 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 ) = 2 x 2 + 5 ⇒ f ( a + r h ) = f ( 1 + r h ) = 2 ( 1 + r h ) 2 + 5 = 2 r 2 h 2 + 4 r h + 7 ∴ I = ∫ 1 3 ( 2 x 2 + 5 ) d x = lim h → 0 h ∑ r = 1 n ( 2 r 2 h 2 + 4 r h + 7 ) = lim h → 0 2 h 3 ∑ r = 1 n r 2 + 4 h 2 ∑ r = 1 n r + 7 h ∑ r = 1 n 1 = lim h → 0 2 h 3 n ( n + 1 ) ( 2 n + 1 ) 6 + 4 h 2 n ( n + 1 ) 2 + 7 h n = lim h → 0 n h ( n h + h ) ( 2 n h + h ) 3 + 2 n h ( n h + h ) + 7 n h = lim h → 0 2 ( 2 + h ) ( 2 × 2 + h ) 3 + 2 × 2 ( 2 + h ) + 7 × 2 { ∵ n h = 2 } = 16 3 + 8 + 14 ∴ I = 82 3

Evaluate ∫ 1 3 ( x 2 + x ) d x as limit of sum.

Let I = ∫ 1 3 ( x 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 = 1 , b = 3 ∴ n h = 3 - 1 = 2 And f ( x ) = x 2 + x ⇒ f ( a + r h ) = f ( 1 + r h ) = ( 1 + r h ) 2 + ( 1 + r h ) = r 2 h 2 + 3 r h + 2 ∴ I = ∫ 1 3 ( x 2 + x ) d x = lim h → 0 h ∑ r = 1 n ( r 2 h 2 + 3 r h + 2 ) = lim h → 0 h 3 ∑ r = 1 n r 2 + 3 h 2 ∑ r = 1 n r + 2 h ∑ r = 1 n 1 = lim h → 0 h 3 n ( n + 1 ) ( 2 n + 1 ) 6 + 3 h 2 n ( n + 1 ) 2 + 2 h n = lim h → 0 n h ( n h + h ) ( 2 n h + h ) 6 + 3 n h ( n h + h ) 2 + 2 n h = lim h → 0 2 ( 2 + h ) ( 2 × 2 + h ) 6 + 3 × 2 ( 2 + h ) 2 + 2 × 2 { ∵ n h = 2 } = 8 3 + 6 + 4 ∴ I = 38 3

Evaluate ∫ 0 2 ( x 2 + x ) d x as limit of sum.

Let I = ∫ 0 2 ( x 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 = 0 , b = 2 ∴ n h = b - a = 2 And f ( x ) = x 2 + x ⇒ f ( a + r h ) = f ( 0 + r h ) = ( r h ) 2 + r h ∴ I = ∫ 0 2 ( x 2 + x ) d x = lim h → 0 h ∑ r = 1 n ( r 2 h 2 + r h ) = lim h → 0 h 3 ∑ r = 1 n r 2 + h 2 ∑ r = 1 n r = lim h → 0 h 3 ( n ) ( n + 1 ) ( 2 n + 1 ) 6 + h 2 n ( n + 1 ) 2 = lim h → 0 n h ( n h + h ) ( 2 n h + h ) 6 + n h ( n h + h ) 2 = lim h → 0 2 ( 2 + h ) ( 2 × 2 + h ) 6 + 2 ( 2 + h ) 2 { ∵ n h = 2 } = 2 × 4 3 + 2 ∴ I = 14 3