Asked by Natalie
Find the points of intersection of the graphs of the functions y=4x^2-25 and y=0. -5/2,5/2
Find the area bounded by the curves y=4x^2-25, y=0, x=-1, x=3.5. PS the area is not 54 I already tried.
Find the area bounded by the curves y=4x^2-25, y=0, x=-1, x=3.5. PS the area is not 54 I already tried.
Answers
Answered by
Reiny
Your intersection points are correct.
The problem is that the second boundary value of x = 3.5 is beyond the intersection point of (2.5,0), and you have a part of the region that lies above the x-axis. So you have to find the area in two parts:
from x = -1 to x=2.5, and then from x = 2.5 to 3.5
area = ∫(25-4x^2) dx from -1 to 2.5 + ∫(4x^2-25)dx from 2.5 to 3.5
= [25x - (4/3)x^3] from -1 to 2.5 + [(4/3)x^3 - 25x] from 2.5 to 3.5
= 62.5 - 125/6 - (-25 + 4/3) + 81/2 - 87.5 - (125/6 - 62.5)
= 60
check my arithmetic
The problem is that the second boundary value of x = 3.5 is beyond the intersection point of (2.5,0), and you have a part of the region that lies above the x-axis. So you have to find the area in two parts:
from x = -1 to x=2.5, and then from x = 2.5 to 3.5
area = ∫(25-4x^2) dx from -1 to 2.5 + ∫(4x^2-25)dx from 2.5 to 3.5
= [25x - (4/3)x^3] from -1 to 2.5 + [(4/3)x^3 - 25x] from 2.5 to 3.5
= 62.5 - 125/6 - (-25 + 4/3) + 81/2 - 87.5 - (125/6 - 62.5)
= 60
check my arithmetic
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