# Evaluate the definite integrals

• Dec 14th 2010, 07:11 AM
watp
Evaluate the definite integrals
I have absolutely no idea how I'm supposed to do this.

Evaluate each of the following definite integrals:

a) $\displaystyle \int_1^4 \frac{x^2+2x}{\sqrt{x^2}} dx$
b) $\displaystyle \int_{-1}^{1} \frac{1}{v}(3v^3-v^\frac{-1}{3}) dv$

Can anyone point me in the right direction? I've done definite integrals, but none that hard before.

• Dec 14th 2010, 09:24 AM
watp
Worked out how to do it.

You integrate it first, then multiply it by the bigger number and minus it by the smaller number times the value.
• Dec 14th 2010, 05:52 PM
Prove It
The easiest way is to simplify the integrand...

In the region $\displaystyle \displaystyle 1 \leq x \leq 4, \sqrt{x^2} = x$.

So $\displaystyle \displaystyle \int_1^4{\frac{x^2+2x}{\sqrt{x^2}}\,dx} = \int_1^4{\frac{x^2+2x}{x}\,dx} = \int_1^4{x + 2\,dx}$.

Also $\displaystyle \displaystyle \int_{-1}^1{\frac{1}{v}(3v^3 - v^{-\frac{1}{3}})\,dv} = \int_{-1}^1{3v^2 - v^{-\frac{4}{3}}\,dv}$.