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$y=\dfrac{1}{\sqrt{36-7x}}\;$ 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

3 Answers 3

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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.

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    Oh! So then it should be $4(\sqrt{30}-3)$ – 2017-02-12
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    Yeah, you're right. – 2017-02-12
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    My answer is actually wrong. I added the constant and the limit and everything. are you sure that should be the right answer? – 2017-02-12
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    There was a minor correction. This is the correct answer. – 2017-02-12
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    @ΘΣΦGenSan I don't get why my answer is wrong. It is the same as the first one. – 2017-02-12
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    @SchrodingersCat I did not down vote. Ah okay, you have the same answer. I looked previously at Kate.K' comment. You deserved an upvote from me.---) – 2017-02-12
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The answer is 1.42

It should be

$\int^0_1(_2)u^(-1/2)*du/7$

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    No explanation, no proper formatting – 2017-02-12
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    Sorry, not sure how to put in the formatting. I always ends up looking wrong. Do you know how to format it? – 2017-02-12
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    Oh Im sorry you answered your own question, forget the explanation part – 2017-02-12
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    Check out my answer, ask if you are unclear about anything – 2017-02-12
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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.