$\int_0^1 \sqrt[3] {1+7x}\; dx $
I have let u = 1+7x and so dx = $\frac 17$du. Then I did when x=0, u=1 and when x=1, u=2. So the integral I am now trying to solve is
$\frac 17\int_1^2 \sqrt[3] {u}\; du $
When I do this I get that the integral of $\sqrt[3] {u}$ is $\frac{3u^\frac43}{4} $.
When I solve for the final value I get $\frac{6\sqrt[3] {2}-3}{28} $ but the answer is suppose to be $\frac {45}{28}$, I'm not sure what I am doing wrong.
When $x=0$, $u=1$. When $x=1$, $u=8$, not $2$. Your upper limit of integration is wrong.