# Thread: Reduction Formula (In terms of x)

1. ## Reduction Formula (In terms of x)

Q: Given that $\displaystyle I_n = \displaystyle\int^8_0 x^n (8-x)^{\frac{1}{3}} \, \mathrm{d}x, \ \ n \ge 0$.
(a) Show that $\displaystyle I_n = \frac{24n}{3n+4}I_{n-1}, \ \ n \ge 1$.
(b) Hence find the exact value of $\displaystyle \displaystyle\int^8_0 x(x+5)(8-x)^{\frac{1}{3}} \, \mathrm{d}x$.
__________________
I've done part (a) and got $\displaystyle I_n = 6nI_{n-1} - \frac{3}{4}nI_n$ but cannot simplify to get it into the form that the question is asking for. Also, I cannot do part (b). Can I have help? Thanks in advance.

2. Originally Posted by Air
Q: Given that $\displaystyle I_n = \displaystyle\int^8_0 x^n (8-x)^{\frac{1}{3}} \, \mathrm{d}x, \ \ n \ge 0$.
(a) Show that $\displaystyle I_n = \frac{24n}{3n+4}I_{n-1}, \ \ n \ge 1$.
(b) Hence find the exact value of $\displaystyle \displaystyle\int^8_0 x(x+5)(8-x)^{\frac{1}{3}} \, \mathrm{d}x$.
__________________
I've done part (a) and got $\displaystyle I_n = 6nI_{n-1} - \frac{3}{4}nI_n$
Then $\displaystyle I_n+\frac 34 n I_n=6n I_{n-1}$

$\displaystyle \frac{4+3n}{4} I_n=6n I_{n-1}$

$\displaystyle I_n=\frac{24n}{3n+4} I_{n-1}$

You were close to it

3. Originally Posted by Air
(b) Hence find the exact value of $\displaystyle \displaystyle\int^8_0 x(x+5)(8-x)^{\frac{1}{3}} \, \mathrm{d}x$.
\displaystyle \begin{aligned} \int^8_0 x(x+5)(8-x)^{\frac{1}{3}} \, \mathrm{d}x &=\int_0^8 (x^2+5x)(8-x)^{\frac 13} dx \\ \\ &=\int_0^8 x^2 (8-x)^{\frac 13} dx+\int_0^8 5x(8-x)^{\frac 13} dx \\ \\ &=I_2+5 \cdot I_1 \end{aligned}

Write $\displaystyle I_2$ with respect to $\displaystyle I_1$