# Thread: fundamental theorem of calculus problem

1. ## fundamental theorem of calculus problem

Find the derivative of the function:
a. $\displaystyle \int$ $\displaystyle \frac{u^2-1}{u^2+1}dx$
Couldn't get the bounds to come out right, but they are 2x to 3x.

b. $\displaystyle \int$ $\displaystyle sqrt(t)*sintdt$
from sqrt(x) to $\displaystyle x^3$.
Can someone help with these?

2. Originally Posted by hazecraze
Find the derivative of the function:
a. $\displaystyle \int$ $\displaystyle \frac{u^2-1}{u^2+1}dx$
Couldn't get the bounds to come out right, but they are 2x to 3x.

b. $\displaystyle \int$ $\displaystyle sqrt(t)*sintdt$
from sqrt(x) to $\displaystyle x^3$.
Can someone help with these?
Both of these are easy if you know the FTC. Part of it says that the derivative of an integral is the original function. When you have bounds from a definite integral you plug in the top bound into f(u) minus the lower bound of f(u). If either bound is not "x" or a constant, you must apply the chain rule and multiply the f(u) by u'.

The Fundamental Theorem of Calculus

3. Originally Posted by hazecraze
Find the derivative of the function:
a. $\displaystyle \int$ $\displaystyle \frac{u^2-1}{u^2+1}dx$
Couldn't get the bounds to come out right, but they are 2x to 3x.

b. $\displaystyle \int$ $\displaystyle sqrt(t)*sintdt$
from sqrt(x) to $\displaystyle x^3$.
Can someone help with these?
$\displaystyle I(x)=\int_{2x}^{3x}\frac{u^2-1}{u^2+1}du$

Use the following properties of integrals:

Split the integral using the property:

$\displaystyle \int_a^bf(x)dx=\int_c^bf(x)dx+\int_a^cf(x)dx$

Use $\displaystyle c=0$

$\displaystyle I(x)=\int_{0}^{3x}\frac{u^2-1}{u^2+1}du+\int_{2x}^0\frac{u^2-1}{u^2+1}du$

You'll also need to use the property:

$\displaystyle \int_a^bf(x)dx=-\int_b^af(x)dx$

To flip the second term.

$\displaystyle I(x)=\int_{0}^{3x}\frac{u^2-1}{u^2+1}du-\int_{0}^{2x}\frac{u^2-1}{u^2+1}du$

Now just differentiate using the chain rule.