SOLUTION: THIS IS INTEGRAL CALCULUS with ANALYTIC GEOMETRY find the area bounded by the curves find points of intersections lower and upper limit the working integral with lower and up

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Question 1019015: THIS IS INTEGRAL CALCULUS with ANALYTIC GEOMETRY
find the area bounded by the curves
find points of intersections
lower and upper limit
the working integral with lower and upper limits
final answer
1 f(x)=4-x^2 and g(x)=x+2
2 f(x)e^x and g(x)=e^-x
3 f(x)= square root of x and g(x)=x3
4 f(x)=x^2 and g(x)=-x^2
thank you so much i will make this a reviewer. Godbless!
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
1) f(x)=4-x^2 and g(x)=x+2
4 - x^2 = x + 2
x^2 +x -2 = 0
(x+2)(x-1) = 0
x = -2 or x = 1
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points of intersection are (-2, 0) and (1, 3)
lower limit is -2 and upper limit is 1
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consider the graph of the two functions f(x) red line, g(x) green line
+graph%28+300%2C+200%2C+-6%2C+5%2C+-10%2C+10%2C+4-x%5E2%2C+x%2B2%29
:
the area bounded by the curves is found by subtracting the area of the curve below from the area of the curve above, in our case
:
we integrate from -2 to 1 (f(x) - g(x))
:
f(x) - g(x) = 4 - x^2 -x - 2 = -x^2 -x + 2
:
now integrate term by term
:
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integral is -x^3/3 -x^2/2 + 2x, evaluate this for x1=-2 and x2=1
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:
(-1/3 -1/2 +2) - (8/3 -4/2 -4) = 7/6 - (-20/6) = 27/6 = 9/2 = 4.5
:
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area is 4.5 square units
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problems 2 - 4 are found by the same method