Integration E(y^2)

[rbcc]

HI,

I'm haveing trouble with this question can some one check this over?

\(\displaystyle \int^{a+b}_{a-b} y^{2} \frac{1}{2b} \dy\)$\displaystyle dy$
.
$\displaystyle \frac{1}{3}y^{3} \frac{1}{2b}]_{a-b}^{a+b}$

$\displaystyle \frac {a^{3}+3a^{2}b+3ab^{2}+b^{3}-a^{3}+3a^{2}b-3ab^{2}+b^{2}}{3}$...$\displaystyle \frac{1}{2b}$
.
$\displaystyle \frac {6a^{2}b+b^{3}}{3}$...$\displaystyle \frac{1}{2b}$
.
$\displaystyle \frac {6a^{2}b+b^{3}}{6b}$
.
$\displaystyle a^{2}+\frac {b^{2}}{6}$

thanks!

 

[galactus]

You almost have it. It should be $\displaystyle a^{2}+\frac{b^{2}}{\boxed{3}}$

 

[BigGlenntheHeavy]

$\displaystyle See \ my \ attachment.$


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[rbcc]

ok i get it now thanks!