Answer to Question #227621 in Calculus for Moe

Question

True or False.

The value of the integral $\displaystyle{\int_{1}^{2}x(x^2-1)^8 dx}$ is an integer

Answer 1

$\displaystyle{\int_{1}^{2}x(x^2-1)^8 dx}$

Let substitute, $u=x^2-1$

Hence, $\frac{du}{dx}=2x,$ $dx=\frac{1}{2x}du$

$=\frac{1}{2}\smallint{u^8du}$

Now solving, $\smallint{u^8du}$

Apply power rule

$\smallint{u^8du}=\frac{u^{n+1}}{n+1}$ With $n=8$

$=\frac{u^9}{18}$

Undo the substitution. $u=x^2-1$

$=\frac{(x^2-1)^9}{18}$

Replacing x with 2 and 1

$=\frac{(2^2-1)^9}{18}-0=1093.5$

False because 1093.5 is not a whole valued number