SOLUTION: Find the area enclosed by the curve, y=25- x^2 and the straight line, y=x+13

Question 813580: Find the area enclosed by the curve, y=25- x^2 and the straight line, y=x+13

Found 2 solutions by richwmiller, Alan3354:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
integral_(-4)^3 (12-x-x^2) dx = 343/6=57.1667
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the area enclosed by the curve, y=25- x^2 and the straight line, y=x+13
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The 2 points of intersection are at x = -4 and x = 3.
y=25- x^2
INT(x) = 25x - x^3/3
INT(3) = 75 - 9 = 66
INT(-4) = -100 + 64/3 = -236/3
Area under parabola to x-axis = 66 + 236/3 = 434/3
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Area under line = 87.5
--> 343/6 sq units
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