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.
 A: 
So enclosed area $\displaystyle = \int_{0}^{2}\left[4x-x^3\right]dx = $
A: your questions asks for the area above the x-axis, that means you just ignore the area under the x-axis that's bound by the line and the curve. 

and it's found by :
$Area = \int_{0}^2 [4x - x^3]dx = [2x^2 - \frac{1}{4}x^4]^2_0 = 2(2)^2 - \frac{1}{4} (2)^4 - (2(0)^2-\frac{1}{4}(0)^4) = 4$  $units^2$ 
the answer in your book is probably wrong, though if you could check with someone that has/knows that book as well that'd be better.
A: 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!
