The Purplemath Forums

by **stripling_warrior** on Thu May 07, 2009 5:54 pm

I have no clue!!!

$A=\int_0^2[(x+6)-x^2]dx$
:?:

**Sponsor**

by **stapel_eliz** on Thu May 07, 2009 7:29 pm

stripling_warrior wrote: $A=\int_0^2[(x+6)-x^2]dx$

Use what you learned back in algebra to expand the polynomial integrand.

Then apply the Power Rule in reverse.

Evaluate the result at x = 2 and at x = 0.

Subtract to obtain the integral value.

If you get stuck, please reply showing how far you have gotten.

Thank you!

by **stripling_warrior** on Thu May 07, 2009 8:35 pm

Thanks!!!!

$=\left[\frac{x^2}{2}+6x-\frac{{x^3}}{3}\right]_0^2= \frac{34}{3}-0= \frac{34}{3}$

The Purplemath Forums

A = integral, from 0 to 2, of (x + 6) - x^2 dx

A = integral, from 0 to 2, of (x + 6) - x^2 dx

Re: A = integral, from 0 to 2, of (x + 6) - x^2 dx