Given vec field $\vec{F}(x,y,z) = xy^3 \vec{i} + x^3y\vec{j}$ find line integral from $(0,0)$ to $(2,8)$ on path $y = 2x^2$ (possible textbook error)

Question

I am give the following problem:

Given the vector field $\vec{F}(x,y,z) = xy^3 \ \vec{i} + x^3y\ \vec{j}$ find the line integral from $(0,0)$ to $(2,8)$ on the path $y = 2x^2$

My answer:

Along the curve we have

\begin{align*} \overrightarrow{F}(x,y,z) &= \left< xy^3, x^3y \right>\\ \vec{r}(t) &= \left< t, 2t^2\right>\\ \\ \vec{F}(\vec{r}(t)) &= \left< 8t^7 , 2t^5 \right> \end{align*}

Calculating the derivative of the parametrization,

\begin{align*} \vec{r}'(t) = \left< 1, 4t \right> \end{align*}

The dot product is given by

\begin{align*} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) &= \left< 8t^7 , 2t^5 \right> \cdot \left< 1, 4t \right>\\ &= 8t^7 + 8t^6 \end{align*}

So the line integral is

\begin{align*} \int_{0}^{1} 8t^7 + 8t^6 \ dt &= \cdots \\ &= \frac{15}{7} \end{align*}

The problem is that my answer differs from the textbook's. Did I make a mistake somewhere?

Textbook's answer: $\frac{2816}{7}$

Thank you.

Answer 1

You integral range is wrong. It should be from 0 to 2 but not from 0 to 1.