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
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$\displaystyle \int_0^1(e^{2x}-e^x)dx=\left.\frac{e^{2x}}{2}\right|_0^1-\left.e^x\right|_0^1=\frac{e^2}{2}-\frac{1}{2}-(e-1)=$ $\displaystyle =\frac{e^2}{2}-e+\frac{1}{2}$
Originally Posted by red_dog $\displaystyle \int_0^1(e^{2x}-e^x)dx=\left.\frac{e^{2x}}{2}\right|_0^1-\left.e^x\right|_0^1=\frac{e^2}{2}-\frac{1}{2}-(e-1)=$ $\displaystyle =\frac{e^2}{2}-e+\frac{1}{2}$ Thanks so much. I realised my mistake.
Last edited by mr fantastic; Sep 7th 2009 at 04:34 AM. Reason: New question moved to new thread.
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