Evaluate the integral. http://www.webassign.net/www21/symIm...ea86047c9c.gif

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- Nov 17th 2007, 02:07 PMwinterwyrmneed to get the concept down...
Evaluate the integral. http://www.webassign.net/www21/symIm...ea86047c9c.gif

- Nov 17th 2007, 02:26 PMbobak
Okay I'm going to have to assume you know how to deal with expression in the form of $\displaystyle \int_{}^{}{ae^{bx}}dx$ If you don't just say.

Now to integrate $\displaystyle 3^x$ we just need to rewrite it as e raised to an particular power, consider $\displaystyle y = 3^x$ then $\displaystyle ln(y) = ln(3^x) .... ln(y) = xln(3)$ now we just antilog both sides so $\displaystyle y = e^{xln3}$

so you understand how to write $\displaystyle 3^x$ as $\displaystyle e^{xln3}$

so to integrate you just use the chain rule backwards $\displaystyle \int_{}^{}{e^{xln3}}dx = \frac{e^{xln3}}{ln3} = \frac{3^{x}}{ln3} $

now in general $\displaystyle \int{a^x} dx = \frac{a^x}{lna}$ - Nov 17th 2007, 03:33 PMwinterwyrm
cool response, but it doesnt work.... F(3)-F(1) isnt' the right answer.....

- Nov 17th 2007, 03:50 PMKrizalid
$\displaystyle (a^x)'=(e^{x\ln a})'=a^x\ln a.$

Integrate both sides

$\displaystyle a^x+k_1=\int a^x\ln a\,dx\,\therefore\,\frac{a^x}{\ln a}+k=\int a^x\,dx.$

So $\displaystyle \int_1^33^x\,dx=\left.\frac{3^x}{\ln3}\right|_1^3$

What's the problem with this? - Nov 17th 2007, 07:26 PMwinterwyrm
I get 21.84 when I plug that in, and when I try to submit it for my online homework..... I get a red X (I have a hundred submissions though)..... is this a serious problem?

- Nov 17th 2007, 08:17 PMJhevon
- Nov 17th 2007, 09:09 PMwinterwyrm
its wrong... 22, 22.9, 21.85, even 21, and 23.... :(

- Nov 17th 2007, 09:16 PMJhevon
- Nov 17th 2007, 09:23 PMwinterwyrm
that worked! THanks so much!! I was so confused!!