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Alex73 [517]
3 years ago
12

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

Mathematics
1 answer:
Soloha48 [4]3 years ago
8 0

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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