Question

How do you evaluate the definite integral `int (x^2-x-2)/(x-2)dx` from [0,1]?

Answer 1

`3/2.`

Explanation:

`I=int_0^1(x^2-x-2)/(x-2)dx=int_0^1{(x-2)(x+1)}/(x-2)dx.`

Since, `x in [0,1], x!=2, :.," so, dividing by "(x-2)ne0,`

we have, `I=int_0^1(x+1)dx,`

`=[x^2/2+x]_0^1.`

`rArr I=3/2.`