What is the correct solution for $\int_{0}^{1}x\sqrt{18-2x^2} \,dx$?

[cbarker1]

Hello,

I need some help finding the integral
$\int_{0}^{1}x\sqrt{18-2x^2} \,dx$
Let $u= 18-2x^2$
$du=-4xdx$
$-1/4 \int_{16}^{18} \sqrt{u} \,du$= $36 sqrt(2) -128/3$

I am getting the wrong solution: $9\sqrt(2)-32/3$

 

[I like Serena]

Hello,

I need some help finding the integral
$\int_{0}^{1}x\sqrt{18-2x^2} \,dx$
Let $u= 18-2x^2$
$du=-4xdx$
$-1/4 \int_{16}^{18} \sqrt{u} \,du$= $36 sqrt(2) -128/3$

I am getting the wrong solution: $9\sqrt(2)-32/3$

Hi Cbarker1! ;)

The intermediate result should be:
$$-1/4 \int_{18}^{16} \sqrt{u} \,du$$
Integrating it, we get:
$$-\frac 14 \int_{18}^{16} \sqrt{u} \,du
=\frac 14 \int_{16}^{18} \sqrt{u} \,du
=\frac 14\cdot \frac 23 u^{\frac 32}\Big|_{16}^{18}
=\frac 16({18}^{\frac 32}-16^{\frac 32})
=\frac 16((3\sqrt 2)^3 -4^3)
=\frac 16(54\sqrt 2-64)
=9 \sqrt 2-\frac{32}3
$$