Question

find the area enclosed by the curve y=x^3-4x,y=0​

Answer 1

Step-by-step explanation:

There are two areas bounded by the curve[math] y=x^3[/math] and line y=4x.

First of all, you need to find where these intersect, that is

[math]x^3=4x\Leftrightarrow x(x^2-4)=0 \Leftrightarrow x=0\vee x=\pm 2 [/math]

If we only look at the area bounded by the curve from 0 to 2 we have that[math] 4x\geq x^3 [/math]between 0 and 2. Thus we have the integral

[math]\int_0^2 4x-x^3 dx = \left[2x^2-\frac{1}{4}x^4\right]_0^2=4[/math]

If you are also supposed to find the area where x<0 you do it similarly.