How to evaluate $$ \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx ~~ ?$$ I have tried to evaluate, $$ I = \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx = 5\int_{0}^2 x^{27} (x-1)^{17} dx - 3\int_{0}^2 x^{26} (x-1)^{17} dx$$ by parts but I am getting very lengthy expressions for both the integrals . Please help.
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6Hint: You may split $(5x-3)=2x+3(x-1)$ and write $I=I_1+I_2$, Integrate $I_1$ such that $-I_2$ appears. – Z Ahmed Jun 26 '19 at 06:29
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@DrZafarAhmedDSc By that way, answer would be zero, not $1.5 \times 10^7$ – xrfxlp Jun 26 '19 at 06:36
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@AjayMishra the answer is not zero. – Pallavi Jun 26 '19 at 06:45
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I know that, answer is latter one, approx. – xrfxlp Jun 26 '19 at 06:45
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That is mechanised in CASes, e.g. Mathematica answers $ \frac{134217728}{9}$ through the command Integrate[x^26(x - 1)^17(5 x - 3), {x, 0, 2}] . – user64494 Jun 26 '19 at 10:16
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Let $$ I = \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx = \int_{0}^2 x^{26} (x-1)^{17} (2x+3(x-1) )dx = I_1 + I_2 $$ Let us integrate $I_1$ by parts, taking $x^{27} $ as first function, therefore, $$ I_1 = \left . \frac{1}{9} x^{27} (x-1)^{18} \right| _{0}^{2}- I_2 ~ = \frac{2^{27}}{9} -I_2 $$ Hence, $$ I = \frac{2^{27}}{9}~~ .$$
Pallavi
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There is a trick, the numbers have been "arranged":
$$(x^{27}(x-1)^{18})'=27x^{26}(x-1)^{17}+18x^{27}(x-1)^{16}=(27(x-1)+18x)x^{26}(x-1)^{17}\\=9(5x-3)x^{26}(x-1)^{17}.$$
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You could feel it by the fact that $27+1+17+1$ and $17+1$ are in the ratio $5:3$. – Jun 26 '19 at 07:01
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$$\frac{d}{dx} x^{a+1} (x-1)^{b+1} = x^a (x-1)^b \left\{(a+b+2)x - (a+1)\right\}$$
So \begin{align*} \int x^{26} (x-1)^{17} \left\{(26+17+2)x - (26+1)\right\} dx &= \int x^{26} (x-1)^{17} (45x - 27) dx \\ &= 9\int x^{26} (x-1)^{17} (5x - 3) dx \\ &= x^{27} (x-1)^{18} \end{align*}
Thus $$\int x^{26} (x-1)^{17} (5x - 3) dx = \frac{1}{9} x^{27} (x-1)^{18}$$
jacky
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