fundamental theorem of calculus problem

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November 30th 2009, 02:13 PM
hazecraze

fundamental theorem of calculus problem

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

b. $\int $ $sqrt(t)*sintdt$
from sqrt(x) to $x^3$ .
Can someone help with these?
November 30th 2009, 02:35 PM
Jameson

Quote:

Originally Posted by hazecraze

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

b. $\int $ $sqrt(t)*sintdt$
from sqrt(x) to $x^3$ .
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'.

See this page for some good examples:

The Fundamental Theorem of Calculus
November 30th 2009, 02:37 PM
adkinsjr

Quote:

Originally Posted by hazecraze

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

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

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

Use the following properties of integrals:

Split the integral using the property:

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

Use $c=0$

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

You'll also need to use the property:

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

To flip the second term.

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

Now just differentiate using the chain rule.