Find k value where the function is a pdf

Question

(a) $kx^6(1-x)^4$, for $0 < x < 1, 0$ otherwise

(b) $kx^2(4-x)^3$, for $0 < x < 4$. $0$ otherwise

my attempt

(a) $$\int_{0}^{1} kx^6(1-x)^4 dx$$

Do I just solve this on $[0,1]?$

The easiest way to solve this is to convert from a .pdf to a .tex. Anyway, getting to the point, you need to integrate it and figure out the precise value of $k$ for which the integral is equal to $1$. So yes your integral is correct. — Harambe, Jun 13 '18 at 03:44
Would there be an easier way to integrate something like this? The only way I can think of is to expand it fully — Bas, Jun 13 '18 at 03:49
it's not hard to expand it fully, it'll take longer trying to find a way to integrate it nicely :) If you're careful you can use the fact that $(1-x)^5$ differentiates to a constant multiple of $(1-x)^4$ but the presence of the $x^6$ term in front makes it difficult to exploit this. — Harambe, Jun 13 '18 at 04:01
you can use easily integration by parts 6 times. It is not hard to figure out the result just with your head. Try it. — Masacroso, Jun 13 '18 at 04:26
Using the binomial theorem, $$ (1-x)^4 = 1 -4x +6x^2 -4x^3 +x^4. $$ — Math1000, Jun 13 '18 at 05:11

score 0 · Accepted Answer · answered Jun 13 '18 at 05:37

Consider the general case of $$I_{m,n}=\int_0^1 x^m (1-x)^n \,dx$$ Let $$x=\sin^2(t) \implies dx=2\sin(t)\cos(t)\,dt$$ making $$I_{m,n}=2 \int_0^{\frac \pi 2} \sin ^{2 m+1}(t) \cos ^{2 n+1}(t)\,dt$$ and using the usual reduction formula $$I_{m,n}=\frac{\Gamma (m+1)\, \Gamma (n+1)}{\Gamma (m+n+2)}$$

Find k value where the function is a pdf

1 Answers1