# Evaluate integral

• May 20th 2009, 08:47 PM
algebra2
Evaluate integral
l don't understand the d/dx in front of the integral...

$\displaystyle \frac{d}{dx}\int^x_0 \frac{du}{1+u^{2}}$

= arctan 0 - arctan x

= -arctanx

could this be right?
• May 20th 2009, 09:24 PM
TheEmptySet
Quote:

Originally Posted by algebra2
l don't understand the d/dx in front of the integral...

$\displaystyle \frac{d}{dx}\int^x_0 \frac{du}{1+u^{2}}$

= arctan 0 - arctan x

= -arctanx

could this be right?

You need to use the fundemental theorem of calculus

$\displaystyle \frac{d}{dx} \int_{0}^{x}\frac{1}{1+u^2}du=\frac{1}{1+x^2}$
• May 20th 2009, 10:33 PM
matheagle
there was no reason to integrate and then differentiate, but you made mistakes in your integration....

Quote:

Originally Posted by algebra2
l don't understand the d/dx in front of the integral...

$\displaystyle \frac{d}{dx}\int^x_0 \frac{du}{1+u^{2}}$

= $\displaystyle {d\over dx} (\arctan x - \arctan 0)$

= $\displaystyle {d\over dx} (\arctan x)$

= $\displaystyle {1\over 1+x^2}$

is right....

.....NOW, but is unnecessary.