Question

How to find Find F' of `int_x^10(t+1/t)dt` Using the Fundamental Theorem of Calulus?

Find F' of `int_x^10(t+1/t)dt` then check the result by first integrating and then differentiating. please help.

Answer 1

`F'=-x-1/x`

Explanation:

The fundamental theorem of calculus states that if `F(x)=int_a^xf(t)dt` then `F'(x)=f(x)`.

So

`F=int_x^10t+1/tdt=-int_10^xt+1/tdt=-x-1/x`

We check this by integrating first and then differentiating

`F(x)=int_x^10t+1/tdt=[t^2/2+ln|t|]_x^10=50+ln10-(x^2/2-ln|x|)=-x^2/2-ln|x|+50+ln10`

So `F'(x)=d/dx(-x^2/2-ln|x|+50+ln10)=-x-1/x`