Question

How do you evaluate the definite integral `int (1/x^2-1/x^3)dx` from [1,2]?

Answer 1

`1/8`.

Explanation:

Since, `intx^ndx=x^(n+1)/(n+1)+C, n!=-1`, we have,

`int_1^2(1/x^2-1/x^3)dx=[x^(-2+1)/(-2+1)-x^(-3+1)/(-3+1)]_1^2`

`=[x^-1/-1-(x^-2/-2)]_1^2=[1/(2x^2)-1/x]_1^2=(1/8-1/2)-(1/2-1)`

`=-3/8-(-1/2)=1/8`.