calculus - How to solve $int^1_0 (1+7x)^{1/3}dx$?

I worked through $\int^1_0 (1+7x)^{1/3}dx$ and I got $\frac{3}{4}+C$ for the answer. However, I forgot about the exponent when I found the difference of the sides of the integral.

I am retrying it, but I've realized I don't know how to find a number to the power of $\frac{4}{3}$. Also, when I went over this with Symbolab, once u-substitution had been applied, the integral changed to $\int^8_1$ for some reason. I'm sure it's the key to solving this, but I have no idea why that's even allowed.

My textbook and Symbolab both say the answer is $\frac{45}{28}$.

Here are the steps I took. Please let me know what I got wrong.

$\int^1_0 (1+7x)^{1/3}dx$

Let $u=1+7x$

Then $du=7dx$ and $dx=\frac{1}{7}du$

so $\int^1_0u^{1/3}\frac{1}{7}du=\frac{1}{7}\int^1_0u^{1/3}$

$$\frac{1}{7}\int^1_0u^{1/3}$$

$$=\frac{1}{7}[\frac{u^{4/3}}{4/3}|^1_0]$$ $$=\frac{1}{7}[\frac{3u^{4/3}}{4}|^1_0]$$

$$=\frac{1}{7}[\frac{3(1+7(1))^{4/3}}{4}-\frac{3(1+7(0))^{4/3}}{4}]$$

$$=\frac{1}{7}[\frac{3(1+7)^{4/3}}{4}-\frac{3(1+0)^{4/3}}{4}]$$

This is as far as I can get.

Answer

You didn't change your limits. When $x=0,$ then $u=1,$ and when $x=1,$ then $u=8.$