Question

Consider the area between the graphs x+3y=14 and x+4=y2

can someone please help calculate i have 5 minutes><<

Answer 1

calculus?

Answer 2

y = (23 - x)/3 and y = ±√(x - 5).
(a) (23 - x)/3 = √(x - 5) ==> x = 14
(b) (23 - x)/3 = -√(x - 5) ==> x = 41.
two regions to consider: one from 5 to 14 and one from 14 to 41
(23 - x)/3 < √(x - 5) on (5, 14) and (23 - x)/3 > -√(x - 5) on (14, 41) area is ∫ [√(x - 5) - (23 - x)/3] dx (from x=5 to 14) + ∫ [(23 - x)/3 + √(x - 5)] dx (from x=14 to 41).
=> a = 5, b = c = 14, d = 41, f(x) = √(x - 5) - (23 - x)/3, and g(x) = (23 - x)/3 + √(x - 5).
from this time you solve for x instead of y and do the same thing as did above. not sure if it is correct.

Answer 3

try integrating the top function minus the bottom function