# Quick calc 1 question

1. Jul 16, 2014

### iScience

$$\int \sqrt{f(x)}dx$$ is NOT the same as..

$$\sqrt{\int f(x)dx}$$ right?

i did an example problem and they turned out not to be equivalent but i just wanted to make certain.

if they actually aren't equivalent.. why aren't they???

2. Jul 16, 2014

### iScience

well, better example:

$$\sqrt{(C\int\sqrt{(a+b)}dx)^2}$$

this is not the same as..

$$\sqrt{C^2\int(a+b)dx}$$

is it?

(C= constant, a= variable, b=constant (not that it really matters..))

Last edited: Jul 16, 2014
3. Jul 16, 2014

### PeroK

Try $f(x)=x^2$ and you'll see for yourself why they are not the same.

4. Jul 16, 2014

### gopher_p

Remember that the definite integral is a kind of sum (technically a limit of sums), and so one way to understand why things may or may not work with the integral is to compare it to a sum.

In this case, if you remember that $\sqrt{a+b}\neq\sqrt{a}+\sqrt{b}$ most of the time, and in general $\sqrt{\sum_1^n a_i}\neq\sum_1^n\sqrt{a_i}$, it's not too hard to see why, most of the time, $\sqrt{\int_a^bf(x)\ dx}\neq\int_a^b\sqrt{f(x)}\ dx$.

5. Jul 16, 2014