[Math] Find the area of the region above the $x$-axis bounded by the line $y=4x$ and the curve $y=x^3$

Question

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.

Answer 1

So enclosed area $\displaystyle = \int_{0}^{2}\left[4x-x^3\right]dx = $