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Integrating an Exponent Help! (1 Viewer)

Thread starter Redyapper
Start date Jun 2, 2013

Redyapper

Member

#1

Hi,

I'm stuck on how to integrate 2e^x/2, between 3 and 0, in exact form.
Please try to show all the steps.

Thanks

Carrotsticks

Retired

#2

$\\ \int_{0}^{3}2e^{2x}dx \\\\= 2 \times \int_{0}^{3}e^{2x} dx \\\\= 2 \times \left [ \frac{e^{2x}}{2} \right ]_0 ^3 \\\\= \left [ e^{2x} \right ]_0 ^3 \\\\= e^6 - e^0 \\\\= e^6 - 1$

P

Parvee

(╯°□°）╯︵ ┻━┻

#3

I think its meant to be 2e^(x/2)
So when you integrate you get 4e^(x/2)
So between 3 and 0 you get 4e(3/2) - 4

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