# Thread: Evaluating Integrals (inverse trig functions)

1. ## Evaluating Integrals (inverse trig functions)

Original Problem:
$
\int_{0}^{1}{4}/{(1+x^2)}dx
$

So I got tan^(-1)x ] 0 to 1<-- supposed to be inverse tanx evaluated from 0 to 1.

Inverse tan of 1 is pi/4 minus inverse tan of 0 is 0, so I end up with pi/4 as the answer. The book says the answer is pi... What am I doing wrong?

2. $\int_0^1 \frac{4}{1+x^2} \, dx = 4\int_0^1 \frac{1}{1+x^2} \, dx = 4\left(\frac{\pi}{4}\right) = \pi$

looks like you forgot about the 4 in the original integral

3. Ah, thanks. I knew that the four would cancel somehow, I just didn't know that you could take it out like that.

Also, what's a derivative of an integral? I know how to find it, but what exactly is it? If I understand correctly, an integral is the area between the x-axis and the function, so what information would the derivative of that yield?

4. Originally Posted by tom ato
Also, what's a derivative of an integral? I know how to find it, but what exactly is it? If I understand correctly, an integral is the area between the x-axis and the function, so what information would the derivative of that yield?
since a definite integral with constant limits of integration yields a constant, its derivative would be zero.

a definite integral with at least one variable limit of integration is another story.

look up the second fundamental theorem of calculus.