# a to the u integral

• September 4th 2013, 06:03 PM
Jason76
a to the u integral
Does look right? http://www.freemathhelp.com/forum/im...s/confused.png As far as correct form, etc..?

$\int_{1}^{5} 3^{2x} dx$.

Using $\int a^{u} du = \dfrac{a^{u}}{\ln a} + C$ and $\dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a$.

$u = 2x$.

$du = 2$.

$a = 3$.

$\dfrac{1}{2} \int_{1}^{5} 3^{u} du$.

$\int_{1}^{5} \dfrac{1}{2} 3^{u} du$.

$\int_{1}^{5} \dfrac{3^{u}}{2} du$.

$\dfrac{3^{2(u)}}{2\ln 3} |_{1}^{5}$.

$\dfrac{3^{2(5)}}{2\ln 3}$ - $\dfrac{3^{2(1)}}{2\ln 3}$.

$\dfrac{3^{10}}{2\ln 3}$ - $\dfrac{3^{2}}{2\ln 3}$.

$\dfrac{59049}{2\ln 3}$ - $\dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}$
• September 4th 2013, 06:30 PM
FelixFelicis28
Re: a to the u integral
Yes, that's correct, but there are a few mistakes that I feel the need to point out.

Quote:

Using $\int a^{u} du = \dfrac{a^{u}}{\ln a} + C$ and $\dfrac{d}{dx} (a^{u}) = a^{u} * du * \ln a$.
This should be $\frac{d}{dx} a^u = \frac{d}{du} \cdot \frac{du}{dx} a^u = a^u \ln a \cdot \frac{du}{dx}$ by the chain rule.

Quote:

$u = 2x$

$du = 2$
This should be $u = 2x \implies \frac{du}{dx} = 2$

Quote:

$\dfrac{3^{2(u)}}{2\ln 3} \bigg|_{1}^{5}$.
I'm not quite sure if this is a typo or not but you have to either re-write your integral after you've integrated in terms of $x$ again if you're going to evaluate it with those limits (which I think you've done by putting the $2$ back in but it got a bit confusing by keeping it in terms of $u$) OR change your limits for $u$ i.e. $u = 2x \implies 1 \to 2, \ 5 \to 10$.

Quote:

$\dfrac{59049}{2\ln 3}$ - $\dfrac{9}{2\ln 3} = \dfrac{59040}{2\ln 3} = \dfrac{29520}{\ln 3}$
That's correct. :)
• September 4th 2013, 11:05 PM
Jason76
Re: a to the u integral
Now, it's all coming together.

Pretty sure, this is right:

$\int_{5}^{1} 3^{2x} dx$.

$u = 2x$.

$du = 2 dx \rightarrow du(\dfrac{1}{2}) = dx \rightarrow \dfrac{du}{2} = dx$.

$\dfrac{1}{2} \int_{5}^{1} 3^{u} du \rightarrow \int_{5}^{1} (\dfrac{1}{2}) 3^{u} du \rightarrow \dfrac{3^{u}}{2}$.

$= \dfrac{3^{u}}{2\ln 3} |_{5}^{1} \rightarrow \dfrac{3^{2x}}{2\ln 3} |_{5}^{1}$.

$\dfrac{3^{2(5)}}{2\ln 3} - \dfrac{3^{2(1)}}{2\ln 3} \rightarrow \dfrac{3^{10}}{2\ln 3} - \dfrac{3^{2}}{2\ln 3}$.

$= \dfrac{59049}{2\ln 3} \rightarrow \dfrac{29520}{\ln 3}$.