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2 years ago

I forgot how to do this please help!!!

Mathematics

1 answer:

pogonyaev2 years ago

7 0

Set up a proportion.

x 5
- = -
5.6 4

4x = 5(5.6)
4x = 28
4x 28
- = -
4 4
x = 7.

Did that help?

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Find the integral, using techniques from this or the previous chapter.<br> ∫x(8-x)3/2 dx

Soloha48 [4]

Answer:

$\int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C$

Step-by-step explanation:

For this case we need to find the following integral:

$\int x(8-x)^{3/2}dx$

And for this case we can use the substitution $u = 8-x$ from here we see that $du = -dx$ , and if we solve for x we got $x = 8-u$ , so then we can rewrite the integral like this:

$\int x(8-x)^{3/2}dx= \int (8-u) u^{3/2} (-du)$

And if we distribute the exponents we have this:

$\int x(8-x)^{3/2}dx= - \int 8 u^{3/2} + \int u^{5/2} du$

Now we can do the integrals one by one:

$\int x(8-x)^{3/2}dx= -8 \frac{u^{5/2}}{\frac{5}{2}} + \frac{u^{7/2}}{\frac{7}{2}} +C$

And reordering the terms we have"

$\int x(8-x)^{3/2}dx= -\frac{16}{5} u^{\frac{5}{2}} +\frac{2}{7} u^{\frac{7}{2}} +C$

And rewriting in terms of x we got:

$\int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C$

And that would be our final answer.

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Answer:

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Step-by-step explanation:

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Hope this makes sense!

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Step-by-step explanation:

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