Find the area below the curve y = (x - 5 )^2 above the x-axis and between the vertical lines x=-2 and x=1.?

Answer 1

A = ∫ x² - 10x + 25 dx ( lims. - 2 and 1)
A = x³/3 - 5x² + 25x
A = (1/3 - 5 + 25) - ( - 8/3 - 20 - 50)
A = 1/3 + 20 + 8/3 + 70
A = 90 + 3
A = 93 units ²

Answer 2

Since the integral geometrically represents area between a curve and the x-axis, we need to evaluate the integral:

int( (x - 5)^2, x = -2 to 1)

using the fundamental theorem of calculus, this is
1/3 * x^3 - 5 * x^2 + 25*x evaluated from -2 to 1, so it's

(1/3 - 5 + 25) - (1/3 * (-2)^3 - 5 * (-2)^2 + 25*(-2)) = 93