# Thread: need to get the concept down...

1. ## need to get the concept down...

Evaluate the integral.

2. 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 $\displaystyle \int{a^x} dx = \frac{a^x}{lna}$

3. cool response, but it doesnt work.... F(3)-F(1) isnt' the right answer.....

4. $\displaystyle (a^x)'=(e^{x\ln a})'=a^x\ln a.$

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?

5. 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?

6. Originally Posted by winterwyrm
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?
can't you enter exact answers? or does the system accept decimal approximations? (it would be 21.85 to two decimal places)

7. its wrong... 22, 22.9, 21.85, even 21, and 23....

8. Originally Posted by winterwyrm
its wrong... 22, 22.9, 21.85, even 21, and 23....
check again to see if you wrote the right question, because i see nothing wrong with the responses given. did you try typing in 24/(ln 3)? this is definately the right answer for the question asked

9. that worked! THanks so much!! I was so confused!!