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Given

$$ I_n = \int_{0}^{2} (2x - x^2)^n dx $$

  1. Compute $I_2$

I simply expanded it into $$ \int_0^2 4x^2 - 4x^3 + x^4 dx $$

and computed it.

  1. Show that

$$ (2n+1)I_n = 2nI_{n-1} $$

I first tried doing integration by parts by writing it as $ \int_0^2 (x)' (2x-x^2)^n dx$ but that led nowhere and I also tried splitting it into $\int_0^2 (2x+x^2)^{n-1} (2x+x^2) dx$ and that didn't work either. How do I do this?

  1. Compute $$ \lim_{n \rightarrow \infty} I_n $$
andreas.vitikan
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3 Answers3

3

We have

$$I_n=\int_0^2(x)'(2x-x^2)^ndx=\underbrace{x(2x-x^2)^n\Bigg|_0^2}_{=0}-2n\int_0^2(x-x^2)(2x-x^2)^{n-1}dx\\=-2n\underbrace{\int_0^2(2x-x^2)(2x-x^2)^{n-1}dx}_{=I_n}+n\underbrace{\int_0^2(2x-2)(2x-x^2)^{n-1}dx}_{=0}+2n\underbrace{\int_0^2(2x-x^2)^{n-1}dx}_{=I_{n-1}}$$

  • How do you get to the second row? How is $ -2n \int_o^2 (...) dx $ equal to that? – andreas.vitikan Feb 18 '15 at 15:25
  • The idea is to add and subtract terms to get the result: I began by adding $x$ to $x-x^2$ so we got $(2x-x^2)$. –  Feb 18 '15 at 15:40
  • I still don't see how – andreas.vitikan Feb 18 '15 at 15:55
  • The first step is to write $$\cdots=-2n\int_0^2(2x-x^2)(2x-x^2)^{n-1}dx+2n\int_0^2 x(2x-x^2)^{n-1}dx$$ then write the last integral on this form $$n\int_0^2(2x-2)(2x-x^2)^{n-1}+2n\int_0^2(2x-x^2)^{n-1}dxdx$$ –  Feb 18 '15 at 16:02
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Alternatively, if you know about the Euler Beta Function, use $x=2u$ and just write $$ I_n =\int_0^2 (2x-x^2)^n dx = 2 \cdot 4^n \int_0^1 (u-u^2)^n du = 2 \cdot 4^n \int_0^1 u^n (1-u)^n du = 2 \cdot 4^n \frac{(n!)^2}{(2n+1)!} $$ It is easy to verify your relation between $I_n$ and $I_{n-1}$ from there.

Asier Calbet
  • 2,480
0

you can rewrite $$I_n = \int_0^2 (2x-x^2)^n\, dx = \int_{-1}^1 (1 - x^2)^n\,dx = 2\int_0^1(1-x^2)^n\, dx = 2\int_0^{\pi/2} \sin^{2n+1}t\, dt$$

if we integrate

$\begin{align} J_n &= \int_0^{\pi/2} \sin^{2n+1}t\, dt \\ &= \int_0^1 \sin^{2n} t \sin t \,dt\\ &= -\sin^{2n} t \cos t |_0^{\pi/2} + 2n\int_0^1 sin^{2n-1} t \cos^2 t \,dt\\ &= 2n\int_0^1 sin^{2n-1} t (1-\sin^2 t )\,dt\\ &= 2nJ_{n-1} - 2nJ_{n} \end{align}$

so we have the recurrence relations $$(2n+1)J_n = 2nJ_{n-1}, (2n+1)I_n = 2nI_{n-1}$$

abel
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