# Exponential Integration

• September 7th 2009, 01:20 AM
Lukybear
Exponential Integration
Show that the area between the curves y=e^x and y=e^(2x) from x=0 to x=1 is given by (1/2e^2 - e +1/2).

Ive tried various times and only get 1/2 e^2 -e
• September 7th 2009, 01:30 AM
red_dog
$\int_0^1(e^{2x}-e^x)dx=\left.\frac{e^{2x}}{2}\right|_0^1-\left.e^x\right|_0^1=\frac{e^2}{2}-\frac{1}{2}-(e-1)=$

$=\frac{e^2}{2}-e+\frac{1}{2}$
• September 7th 2009, 01:52 AM
Lukybear
Quote:

Originally Posted by red_dog
$\int_0^1(e^{2x}-e^x)dx=\left.\frac{e^{2x}}{2}\right|_0^1-\left.e^x\right|_0^1=\frac{e^2}{2}-\frac{1}{2}-(e-1)=$

$=\frac{e^2}{2}-e+\frac{1}{2}$

Thanks so much. I realised my mistake.