Find the area of the region R that lies under the graph of $y=\dfrac{1}{\sqrt{36-7x}}\;$ on the interval  $−12  \le  x  \le  0$
I know I use integration by parts and that $u= 36-7x$ and $du=-7$
but I can't find find out how du=dx and how to solve the equation from there. I know something should be in front of the du but I don't know what
 A: The answer is 1.42
It should be 
$\int^0_1(_2)u^(-1/2)*du/7$
A: The area under the curve is equal to: 
$$\int^{0}_{-12} \frac{1}{\sqrt{36-7x}}dx$$
Let $u = 36-7x$
$du = -7 dx$
$\frac{-du}{7} = dx$
Thus our integral now becomes:
$$-\frac{1}{7}\int^{36}_{120} \frac{1}{\sqrt{u}}du$$
How the $36$ and $120$?
Remember, since we are writing everything in terms of $u$ now, we have $u=36-7x$.
So, $36-7(0) = 36$
$36-7(-12) = 120$
Now it's easy.
$$-\frac{1}{7}\int^{36}_{120} (u)^{-1/2}du$$
$$\frac{(u)^{1/2}}{\frac{1}{2}} \text{ Performed power rule}$$
$$2\sqrt{u} \text{ Evaluate at limits}$$
$$-\frac{1}{7}(2\sqrt{36} - 2\sqrt{120}) = 1.41555747146 = 1.42$$
Please ask if you do not understand.
A: The area will be given by the integral $$\int y \, \mathrm{dx}$$
Make the substitution $36-7x=u^2$ and then you will have $-7 \mathrm{dx} = 2u \mathrm{du}$.
And the limits will be changed to:
$\sqrt{120} \ge u \ge 6$
The integral will then be:
$$\frac17 \int_6^\sqrt{120} \frac{2u}{u}du$$
Hope you can solve this now.
