# Integral of (x-7)(3-2x)^(3/2)

• Oct 25th 2009, 01:08 PM
emonimous
Integral of (x-7)(3-2x)^(3/2)
Hey guys, I have this problem I have been trying to solve for the longest time.
Not looking for a complete solution, maybe part of one would be nice.

I'm trying to integrate $\int(x-7)\sqrt{3-2x}^3$

So far I have tried 2 ways of going at it.

First I completely expanded the integral

$\int(x-7)\sqrt{3-2x}^3=\int(-2x^2)(\sqrt{3-2x})+17x\sqrt{3-2x}-21\sqrt{3-2x}$

I have no problem integrating
$-21\sqrt{3-2x}$

I just simply use the substitution technique.
$u=3-2x$

The next part gets tricky, I'm stumped. Any of the functions with x variables I simply can't seem to do.
I tried substituting by parts by either making $u=x$ $dv=3-2x$ or $dv=\sqrt{3-2x}$ so on and so forth
I went to wolframalpha.com and tried to understand what they did step by step

For example, I tried to integrate $\int -21x^2\sqrt{3-2x}$ just like the site did. They say to substitute $u=\sqrt{3-2x}$ and $du=1/\sqrt{3-2x}$ to get

$2\int(1/4)u^2(3-2u^2)^2du$

Either I am feeling really dumb right now and the answer is right in from of me or the answer is completely wrong. How did wolframalpha manage to get
$\int -21x^2\sqrt{3-2x}$= $2\int(1/4)u^2(3-2u^2)^2du$

All help is greatly appreciated
• Oct 25th 2009, 01:39 PM
utopiaNow
Quote:

Originally Posted by emonimous
I'm trying to integrate $\int(x-7)\sqrt{3-2x}^3 \,dx$

From that point use $s = 3 - 2x$, then $ds = -2dx$.
I'll let you crunch through the integral as it should be pretty straightforward. But you should get $\frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$, where C is a constant.
• Oct 27th 2009, 01:09 PM
emonimous
Oh wow! Thanks a bunch!
• Oct 27th 2009, 02:11 PM
emonimous
Quote:

Originally Posted by utopiaNow
From that point use $s = 3 - 2x$, then $ds = -2dx$.
I'll let you crunch through the integral as it should be pretty straightforward. But you should get $\frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$, where C is a constant.

Actually when I try to find the derivative of $\frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$ I get
$(12x-18)(3 - 2x)^{5/2}$
• Oct 27th 2009, 02:23 PM
utopiaNow
Quote:

Originally Posted by emonimous
Actually when I try to find the derivative of $\frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$ I get
$(12x-18)(3 - 2x)^{5/2}$

Well that's incorrect, when I calculate the derivative of $\frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$ I get $(x - 7)(3 - 2x)^{3/2}$, which was precisely the original function whose integral we calculated. Perhaps you should try posting your steps here so I can spot where you went wrong.
• Oct 27th 2009, 02:46 PM
emonimous
Quote:

Originally Posted by utopiaNow
Well that's incorrect, when I calculate the derivative of $\frac{1}{14}(3 - 2x)^{7/2} + \frac{11}{10}(3 - 2x)^{5/2} + C$ I get $(x - 7)(3 - 2x)^{3/2}$, which was precisely the original function whose integral we calculated. Perhaps you should try posting your steps here so I can spot where you went wrong.

Oh don't get me wrong, I was trying to follow your instructions but I think I failed miserably. So I tried to back trace what you did by finding it's derivative.
First I seperated the function in 2 functions so I can derive each
$d/dx \frac{1}{14}(3 - 2x)^{7/2} =\frac{1}{14}[\frac{-7}{2}(3 - 2x)^{5/2}*-2]=\frac{-1}{2}(3 - 2x)^{5/2}$

$
d/dx \frac{11}{10}(3 - 2x)^{5/2}=\frac{-11}{2}[(3 - 2x)^{3/2}]
$

Combining them I get
$\frac{-1}{2}(3 - 2x)^{5/2}+\frac{-11}{2}[(3 - 2x)^{3/2}$
$
= (3 - 2x)^{3/2}(\frac{-11}{2}-\frac{1}{2}(3-2x))
=(3 - 2x)^{3/2}(-6(3-2x))
=(3 - 2x)^{3/2}(-18+12x)
$
• Oct 27th 2009, 02:46 PM
emonimous
Combining them I get
$\frac{-1}{2}(3 - 2x)^{5/2}+\frac{-11}{2}[(3 - 2x)^{3/2}= (3 - 2x)^{3/2}(\frac{-11}{2}-\frac{1}{2}(3-2x))=(3 - 2x)^{3/2}(-6(3-2x))
=(3 - 2x)^{3/2}(-18+12x)

$
• Oct 27th 2009, 03:16 PM
utopiaNow
Quote:

Originally Posted by emonimous
Oh don't get me wrong, I was trying to follow your instructions but I think I failed miserably. So I tried to back trace what you did by finding it's derivative.
First I seperated the function in 2 functions so I can derive each
$d/dx \frac{1}{14}(3 - 2x)^{7/2} =\frac{1}{14}[\frac{-7}{2}(3 - 2x)^{5/2}*-2]=\frac{-1}{2}(3 - 2x)^{5/2}$

$
d/dx \frac{11}{10}(3 - 2x)^{5/2}=\frac{-11}{2}[(3 - 2x)^{3/2}]
$

Combining them I get
$\frac{-1}{2}(3 - 2x)^{5/2}+\frac{-11}{2}[(3 - 2x)^{3/2}$
$
= (3 - 2x)^{3/2}(\frac{-11}{2}-\frac{1}{2}(3-2x))
$

$
=\color{red}(3 - 2x)^{3/2}(-6(3-2x))
=\color{black}(3 - 2x)^{3/2}(-18+12x)
$

The problem is in the red step and obviously the step that follows from it. You see you did $\frac{-11}{2} - \frac{1}{2} = -6$. Which would be fine if that's what it was, but you see the $\frac{-1}{2}$ factor is attached to a $(3 - 2x)$ term, hence what you did does not make any sense. If you further factor out the $(-1/2)$ and then do $(11 + (3 - 2x))$ and carry forwards you will get the correct answer.
• Oct 27th 2009, 03:53 PM
utopiaNow
I'll do a step-by-step solution if you're having difficulty with the integral. Or just to give you something to compare your work to.

Let $I = \int(x-7)(3 - 2x)^{3/2} \,dx$

Then, let $s = 3 - 2x$, hence $ds = -2dx$ ie. $\left(\frac{-1}{2}\right)ds = dx$.

Then $x = \frac{3 - s}{2}$ Hence $x - 7 = \frac{-11 - s}{2}$

Therefore $I = \frac{-1}{2}\int(11 + s)s^{3/2}\left(\frac{-1}{2}\right)ds$

So by multiplying out we get $\frac{1}{4}\int 11s^{3/2} + s^{5/2} = \frac{1}{4}\left(\frac{22}{5}s^{5/2} + \frac{2}{7}s^{7/2}\right) + C$, since $s = 3 - 2x$ we get

$I = \frac{11}{10}(3 - 2x)^{5/2} + \frac{1}{14}(3 - 2x)^{7/2} + C$, which we can actually further factor to give $\frac{-1}{35}(3 - 2x)^{5/2}(5x - 46) + C$ where $C \in \mathbb{R}$ of course.