Answer
$$\int_0^1\sqrt[3]{1+7x}dx=\frac{45}{28}$$
Work Step by Step
To evaluate the integral $$\int_0^1\sqrt[3]{1+7x}dx$$
we will use substitution $1+7x=t$ which gives us $7dx=dt\Rightarrow dx=\frac{dt}{7}$ and the integration bounds would be: for $x=0$ we have $t=1$ and for $x=1$ we have $t=8.$ Now, putting all this into the integral we have:
$$\int_0^1\sqrt[3]{1+7x}dx=\int_1^8\sqrt[3] t\frac{dt}{7}=\frac{1}{7}\int_1^8t^{1/3}dt=\frac{1}{7}\left(\left.\frac{t^{4/3}}{\frac{4}{3}}\right|_1^8\right)=\frac{1}{7}\frac{3}{4}(8^{4/3}-1^{4/3})=\frac{3}{28}(\sqrt[3]{8^4}-1)=\frac{3}{28}(16-1)=\frac{45}{28}$$
