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Alona [7]
3 years ago
11

∫x(x-6)³dx by using substitution u=x-6​

Mathematics
1 answer:
Agata [3.3K]3 years ago
7 0

Answer:

\large\boxed{\int\bigg(x(x-6)^3\bigg)dx=\dfrac{1}{5}(x-6)^5+\dfrac{3}{2}(x-6)^4+C}

Step-by-step explanation:

\int\bigg(x(x-6)^3\bigg)dx\Rightarrow\left[\begin{array}{ccc}x-6=u\\x=u+6\\dx=du\end{array}\right]\Rightarrow\int\bigg((u+6)u^3\bigg)du\\\\=\int(u^4+6u^3)du=\dfrac{1}{5}u^5+\dfrac{6}{4}u^4+C=\dfrac{1}{5}(x-6)^5+\dfrac{3}{2}(x-6)^4+C

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Step-by-step explanation:

1. the measure for HJ you can already see, start from h and draw a line to J. The measure shows 63 degrees.

2. start at F and draw a line to G first. the measure shows 65 degrees. Now continue the line from where you stopped at G, to H. There's no measure but you can see it is in a semicircle. A semicircle is 180 degrees andthe other two angles are 63 and 65.

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3. The meaure is CDE. CD as you can see is a right angles, so 90 degrees. But there is no measure for DE. If you look to the angle vertical from DE which is BA. It measures 40 degrees. DE and BA are vertical so they are congruent. If DE equals 40  and CD equals 90, put them together and you get 130.

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LNP=225

hope this helps

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