<PRE><B>Question 1019015</B>
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( 300, 200, -6, 5, -10, 10, 4-x^2, x+2)}}}
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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
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we integrate from -2 to 1 (f(x) - g(x))
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f(x) - g(x) = 4 - x^2 -x - 2 = -x^2 -x + 2
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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</PRE>