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I want to find the integral of $x (1-x)^8$. How do I go about this? For example, which rule do I use from http://integral-table.com ? Thanks!

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    $\begingroup$ $u$ substitution: let $u=1-x$ (and some algebra after you write the integral in terms of $u$). $\endgroup$ Nov 12, 2011 at 18:26
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    $\begingroup$ If you really want to use that table, then you could try (7). Note that $(1 - x)^8 = (x - 1)^8$, as $(-1)^8 = 1$. $\endgroup$ Nov 12, 2011 at 18:40

4 Answers 4

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$$\begin{align*} \int x(1-x)^8\,dx &=\int x(x-1)^8\,dx\\ &=\int (x-1+1)(x-1)^8\,dx\\ &=\int ((x-1)^9+(x-1)^8)\,dx\\ &=\frac1{10}(x-1)^{10}+\frac19(x-1)^9+C\quad (C: \text{constant}). \end{align*}$$

Note that in the last line I used $$\int (x+a)^n\,dx=\frac1{n+1}(x+a)^{n+1}+C,\quad \text{when}\ n\ne -1.$$

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    $\begingroup$ why do you go from first step to the 2nd step? the part where you used (x-1+1). and what rule did you use from 2nd to 3rd step? PS: I have the solutions, I just need to understand why it is done this way. $\endgroup$
    – nubela
    Nov 12, 2011 at 18:34
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    $\begingroup$ $(x-1+1)(x-1)^8=(x-1)\cdot (x-1)^8+1\cdot (x-1)^8$ follows from the distributive law. $\endgroup$
    – pharmine
    Nov 12, 2011 at 18:40
  • $\begingroup$ Why $x=(x-1)+1$? Because I want to treat $x-1$ as sort of a unit, which makes the calculation easier. $\endgroup$
    – pharmine
    Nov 12, 2011 at 18:43
  • $\begingroup$ That's very slick... I wish i could do +2 $\endgroup$ Nov 12, 2011 at 18:48
  • $\begingroup$ sleek indeed. nice. $\endgroup$
    – nubela
    Nov 12, 2011 at 19:02
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Nobody seems to have mentioned what I'd have thought was an obvious point: It's simpler if the thing raised to the 8th power is just a variable rather than $1$ minus a variable, so let $$ \begin{align} u & = 1-x, \\ \\ du & = - dx, \\ \\ x & = 1-u. \end{align} $$ Then $$ \int x (1-x)^8\;dx = \int (1-u)u^8\;(-du) = \int u^9-u^8\;du. $$ It's easy to antidifferentiate that, and then put $1-x$ wherever $u$ appears.

(After that, the 10th and 9th powers of $1-x$ can be expanded if desired.)

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    $\begingroup$ Someone mentioned it in a comment on the question, but it's nice to have an answer. $\endgroup$ Nov 12, 2011 at 19:25
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    $\begingroup$ The remark is also apparent in a previous answer. $\endgroup$
    – Did
    Nov 12, 2011 at 19:46
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  • Method $1$: Use the binomial expansion of $(1-x)^{8}$ and then integrate term by term.

  • Method $2$: Use integration by parts, by taking $u=x$ and $dv = (1-x)^{8} \ dx$.

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$$x(1-x)^8 = x-8 x^2+28 x^3-56 x^4+70 x^5-56 x^6+28 x^7-8 x^8+x^9$$

so

$$\begin{array}{cl} && \int x(1-x)^8 \,dx \\ &=& \int x-8 x^2+28 x^3-56 x^4+70 x^5-56 x^6+28 x^7-8 x^8+x^9 \,dx \\\\ &=& \int x \,dx - 8 \int x^2 \,dx +28 \int x^3 \,dx - 56 \int x^4 \,dx + 70 \int x^5 \,dx - 56 \int x^6 \,dx + 28 \int x^7 \,dx - 8 \int x^8 \,dx + \int x^9 \,dx \\ &=& \tfrac{1}{2} x^2 - \tfrac{8}{3} x^3 + \tfrac{28}{4} x^4 - \tfrac{56}{5}x^5 + \tfrac{70}{6} x^6 - \tfrac{56}{7} x^7 + \tfrac{28}{8} x^8 - 8 \tfrac{1}{9} x^9 + \tfrac{1}{10} x^{10} + C \end{array}$$

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    $\begingroup$ Eegah... $\endgroup$
    – Did
    Nov 12, 2011 at 20:07
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    $\begingroup$ +1; nice to have a demonstration that the problem does not require cleverness. Finding a slicker solution is just a bonus. $\endgroup$ Nov 12, 2011 at 20:56
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    $\begingroup$ @Henning, sure, so let us attack $x(1-x)^{42}$ now. No cleverness please. Heigh-ho... $\endgroup$
    – Did
    Nov 12, 2011 at 21:33
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    $\begingroup$ No cleverness, but I'm afraid this margin is too small all the same. $\endgroup$ Nov 12, 2011 at 21:40

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