Question

1. Find the area bounded by the curves y= x^2-4, y=0 and x=4.

Answer 1

If we set y = 0 we see that x^2 − 4 = 0, and so x = 2 or x = -2. Thus, the curve cuts the x-axis at x = -2 and at x = 2.

The required area A is entirely above the x-axis and so we can simply evaluate the following definite integral

$A=\int\limits_{2}^{4}(x^2-4)dx={[\frac{x^3}{3}-4x] }_{2}^4=\\ =[\frac{4^3}{3}-4\cdot 4]-[\frac{2^3}{3}-4\cdot 2]=\\=\frac{64}{3}-16-\frac{8}{3}+8=\frac{32}{3}$ .

Answer: $A=\frac{32}{3}$