Find the area of the region above the $x$-axis bounded by the line $y=4x$ and the curve $y=x^3$? Find the area of the region above the $x$-axis bounded by the line $y=4x$ and the curve $y=x^3$
Attempt: 
intersect when:- 
$x^3 - 4x = 0$ 
$x ( x² - 4 ) = 0 $
$x = 0 , x = \pm 2$ 
Area is given by :- 
$$2 ∫_0^2 4x - x^3\; dx = 2 [ 2x^2 - \frac13x^3 ]_0^2$$ 
$$2 [ 8 - 8/3 ] = 32/3\;\;\text{units²}$$
I want to understand this topic well so I'm solving different questions from textbooks under it. The answer in the textbook for this is 6.75 units². I can't figure where I'm wrong.

The real problem is that $\displaystyle \int x^3 = \frac{x^4}{4}$, not $\displaystyle  \frac{x^3}{3}$.
$$\int_0^2 4x-x^3$$
$$= \left[4 \cdot \frac{x^2}{2} - \frac{x^\color{red}{4}}{\color{red}{4}} \right]_0^2$$
$$= \left[ 2x^2  - \frac{x^4}{4} \right]_0^2$$
$$= \left(2\cdot 2^2 - \frac{2^4}{4} \right) - \left(2\cdot 0^2 - \frac{0^4}{4} \right)$$
$$ = (8 - 4) - (0)$$
$$ = 4$$
This agrees with the answer given by Wolfram Alpha, so your textbook is probably wrong. This shows that even textbooks make mistakes!
