Question

Evaluate: `int_1^2 [1/x - 1/(2x^2)]e^(2x) dx`

Answer 1

`int_1^2 [1/x - 1/(2x^2)]e^(2x) dx`

Put 2x = y

⇒ 2dx = dy

⇒ `dx = dy/2`

For x = 1, y = 2 and for x = 2, y = 4

= `1/2 int_2^4 (2/y - 2/y^2)e^y dy`

= `int_2^4 (1/y - 1/y^2)e^y dy`

= `[1/y e^y]_2^4`

= `1/4 e^4 - 1/2 e^2`

= `e^2/2 [e^2/2 - 1]`