Question

) Solve the area bounded by the curve y = 4x-x² and the lines x = -2 and y = 4.

Answer 1

The area of the region bounded by the curves equals a definite integral between the points of intersection of these curves: -2 and 2:

$A=\int_{-2}^2(y_2-y_1)dx=\int_{-2}^2(4-4x+x^2)dx=$

$=(4x-4\frac{x^2}{2}+\frac{x^3}{3})|_{-2}^2=(4x-2x^2+\frac{x^3}{3})|_{-2}^2=$

$=(4\cdot2-2\cdot2^2+\frac{2^3}{3})-(4\cdot(-2)-2\cdot(-2)^2+\frac{(-2)^3}{3})=$

$=(8-8+\frac{8}{3})-(-8-8-\frac{8}{3})=\frac{8}{3}+8+8+\frac{8}{3}=$

$=16+\frac{16}{3}=16+5\frac{1}{3}=21\frac{1}{3}.$

Answer: $21\frac{1}{3}.$