Question
Find the area enclosed between the straight line y=12x+14 and the curve y=3x^2+6x+5
Answers
oobleck
The graphs intersect at (-1,2) and (3,50)
So, consider the area as a collection of thin strips, whose height is the distance between the curves. That makes the area
∫[-1,3] (12x+14)-(3x^2+6x+5) dx = ∫[-1,3] -3x^2 + 6x + 9 dx
= 3∫[-1,3] -x^2 + 2x + 3 dx = 32
So, consider the area as a collection of thin strips, whose height is the distance between the curves. That makes the area
∫[-1,3] (12x+14)-(3x^2+6x+5) dx = ∫[-1,3] -3x^2 + 6x + 9 dx
= 3∫[-1,3] -x^2 + 2x + 3 dx = 32