$I_n=\int_0^\frac{1}{2}(1-2x)^ne^xdx$
Prove that for $n\ge1$ $$I_n=2nI_{n-1}-1$$
I end up (by integrating by parts) with: $I_n =e^x(1-2x)^n+2nI_{n-1}$
I am not sure how $e^x(1-2x)^n$ becomes $-1$?
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HINT: $$\int(1-2x)^ne^xdx$$
$$=(1-2x)^n\int e^xdx-\int\left(\frac{(1-2x)^n}{dx}\int e^xdx\right)dx$$
$$=(1-2x)^ne^x-\int\left(n(1-2x)^{n-1}(-2)e^x\right)dx$$
$$=(1-2x)^ne^x+2n\int (1-2x)^{n-1}e^x dx$$
$$\implies \int_0^{\frac12}(1-2x)^ne^xdx=\left[(1-2x)^ne^x\right]_0^{\frac12}+\int_0^{\frac12} (1-2x)^{n-1}e^x dx$$
$$\text{Now, }(1-2x)^ne^x\big|_0^{\frac12}=0-1=-1$$
lab bhattacharjee
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I dont think you can apply integration by parts in case of indefinite integrals. – Abhra Abir Kundu Jun 12 '13 at 09:09
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@AbhraAbirKundu, Is integration by parts applicable only for definite integral? http://en.wikipedia.org/wiki/Integration_by_parts – lab bhattacharjee Jun 12 '13 at 09:14
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I am not sure..... yet i am saying that look at the derivation they have used the fact that they are integrating it over an interval..... – Abhra Abir Kundu Jun 12 '13 at 09:16
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Moreover the proof I know also integrates it over an interval and then uses the fundamental theorem of calculus. It would appretiate if you can prove it for the indefinite case using the definite case. – Abhra Abir Kundu Jun 12 '13 at 09:17
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@AbhraAbirKundu, please substantiate "I dont think you can apply integration by parts in case of indefinite integrals" – lab bhattacharjee Jun 12 '13 at 09:41
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$\displaystyle I_n=\int_0^\frac{1}{2}(1-2x)^ne^xdx$
$=\displaystyle [(1-2x)^n\int e^{x}dx]_{0}^{1/2}-\int_{0}^{1/2}((\int e^xdx) (\frac{d}{dx}(1-2x)^{n}))dx$
$\displaystyle =[(1-2x)^ne^{x}]_{0}^{1/2}-\int_{0}^{1/2}e^xn(-2)(1-2x)^{n-1}dx$
$\displaystyle =[(1-2x)^ne^{x}]_{0}^{1/2}+2n\int_{0}^{1/2}e^x(1-2x)^{n-1}dx$
$\displaystyle =-1+2nI_n$
$[(1-2x)^ne^{x}]_{0}^{1/2}=(1-2.\frac{1}{2})^ne^{1/2}-(1-2.0)e^0=-1$
Abhra Abir Kundu
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When you apply $x=1/2$ you get $(1-2\frac{1}{2})^n\exp(\frac{1}{2})=0$, while applying $x=0$ you get $(1-0)^n\exp(0)=1$, as $1^n=1$ and $\exp(0)=1$. The difference between the two contributions is then equal to $-1$. – Avitus Jun 12 '13 at 09:19
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