# Definite Integral Problem-Work needs checked

• Apr 12th 2009, 08:15 PM
Jim Marnell
Definite Integral Problem-Work needs checked
$\displaystyle \int_{x=0}^{x=1} (x^2+e^x-6)dx$

$\displaystyle \int (x^2+e^x-6)dx$

$\displaystyle \frac{x^3}{3}+e^x-6x+c$

$\displaystyle \int_{x=0}^{x=1} (x^2+e^x-6)dx$

$\displaystyle \frac{x^3}{3}+e^x-6x\int_{x=0}^{x=1}$

$\displaystyle \frac{1^3}{3}+e^1-6(1)$

$\displaystyle =-5\frac{2}{3}+e$

$\displaystyle \frac{0^3}{3}+e^0-6(0)$

$\displaystyle =e$

$\displaystyle -5\frac{2}{3}+e-e$

$\displaystyle =-5\frac{2}{3}$
• Apr 12th 2009, 08:28 PM
mollymcf2009
Quote:

Originally Posted by Jim Marnell
$\displaystyle \int_{x=0}^{x=1} (x^2+e^x-6)dx$

$\displaystyle \int (x^2+e^x-6)dx$

$\displaystyle \frac{x^3}{3}+e^x-6x+c$

$\displaystyle \int_{x=0}^{x=1} (x^2+e^x-6)dx$

$\displaystyle \frac{x^3}{3}+e^x-6x\int_{x=0}^{x=1}$

$\displaystyle \frac{1^3}{3}+e^1-6(1)$

$\displaystyle =-5\frac{2}{3}+e$

$\displaystyle \frac{0^3}{3}+e^0-6(0)$

$\displaystyle =e$

$\displaystyle -5\frac{2}{3}+e-e$

$\displaystyle =-5\frac{2}{3}$

$\displaystyle e^0 \neq e$
$\displaystyle e^0 = 1$