Consider the area between the graphs x+3y=3 and x+7=y^2. This area can be computed in two different ways using

Consider the area between the graphs x+3y=3 and x+7=y^2. This area can be computed in two different ways using integrals

First of all it can be computed as a sum of two integrals

*integral from a to b* ∫ f(x) dx + *integral from b to c* ∫ g(x) dx

where a = ? , b = ?, c = ? and f(x) = ? , g(x) =?

x + 3y = 3 and x + 7 = y²

Sketch the graphs. You will note that the nose of the sideways parabola contains part of the area while the rest of the area is dominated by the linear function.

We need to know the x-coordinates of the intersection points of the two functions.

Solve simultaneously for y:
x + 7 = y²
x + 3y = 3→ 7 – 3y = y² – 3 → 0 = y² + 3y – 10 = (y + 5)(y – 2)

So y = -5 or y = 2
Corresponding x: x = 25 – 7 = 18 or x = 4 – 7 = -3

Area = ∫[-7 to -3] {√(x + 7) – -√(x + 7)}dx + ∫[-3 to 18] {(3 – x)/3 – -√(x + 7)}dx
= ∫[-7 to -3] {2√(x + 7)}dx + ∫[-3 to 18] {(3 – x)/3 + √(x + 7)}dx
= 32/3 + 93/2 = 57.1666…