Evaluate ∫ 0 1 1 x 2 + 1 d x .

Given integral is ∫ 0 1 1 1 + x 2 d x = tan - 1 x 0 1 ∵ ∫ 1 1 + x 2 d x = tan - 1 x + C = tan - 1 1 - tan - 1 0 = π 4 - 0 = π 4 Hence, ∫ 0 1 1 1 + x 2 d x = π 4

Evaluate: ∫ 0 1 2 π x 2 sin x d x .

Let I = ∫ 0 1 2 π x 2 sin x d x ⇒ I = x 2 ∫ 0 1 2 π sin x d x - ∫ 0 1 2 π d d x x 2 ∫ 0 1 2 π sin x d x d x {Using integration by parts method} ⇒ I = - x 2 cos x 0 1 2 π - ∫ 0 1 2 π 2 x - cos x d x ⇒ I = - x 2 cos x 0 1 2 π + 2 x ∫ 0 1 2 π cos x d x - ∫ 0 1 2 π d d x x ∫ 0 1 2 π cos x d x d x {Using integration by parts method} ⇒ I = - x 2 cos x 0 1 2 π + 2 x sin x + cos x 0 1 2 π ⇒ I = - π 2 2 cos π 2 - 0 + 2 π 2 sin π 2 + cos π 2 - cos 0 ⇒ I = 2 π 2 - 1 ⇒ I = π - 2 Thus, the required value of ∫ 0 1 2 π x 2 sin x d x = π - 2 .

Show that ∫ 0 1 ( x + 2 ) e - 2 x d x = 5 4 - 7 4 e 2 .

To prove: ∫ 0 1 ( x + 2 ) e - 2 x d x = 5 4 - 7 4 e 2 Let I = ∫ 0 1 ( x + 2 ) e - 2 x d x ⇒ I = ∫ 0 1 x e - 2 x d x + ∫ 0 1 2 e - 2 x d x ⇒ I = x ∫ 0 1 e - 2 x d x - ∫ 0 1 d d x x ∫ 0 1 e - 2 x d x d x + 2 ∫ 0 1 e - 2 x d x {Using integration by parts method} ⇒ I = x e - 2 x - 2 0 1 - ∫ 0 1 e - 2 x - 2 d x + 2 e - 2 x - 2 0 1 ⇒ I = x e - 2 x - 2 0 1 + e - 2 x - 4 0 1 + 2 e - 2 x - 2 0 1 ⇒ I = e - 2 - 2 + e - 2 - 4 - e 0 - 4 + 2 e - 2 - 2 - e 0 - 2 ⇒ I = - 1 2 e 2 - 1 4 e 2 + 1 4 - 1 e 2 + 1 ⇒ I = 5 4 - 7 4 e 2 Thus, ∫ 0 1 ( x + 2 ) e - 2 x d x = 5 4 - 7 4 e 2 .