Thursday

September 3, 2015
Posted by **Ashley** on Wednesday, November 28, 2012 at 6:48pm.

the integral from 0 to 5 of (2^x +8)dx.

I got 40+ 31/log(2)but that's wrong.

- Calculus -
**Steve**, Wednesday, November 28, 2012 at 6:59pm∫2^x + 8 dx = 1/ln2 2^x + 8x

so, using the limits, we have

(2^5/ln2 + 40) - (2^0/ln2 + 0)

= 32/ln2 + 40 - 1/ln2

= 31/ln2 + 40

you are correct.