Question 1075206
Same problem, different numbers.
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Use a graphing utility, where helpful, to find the area of the region enclosed by the curves. 
 y=x^3-13x^2+30x and y=0.
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 y=x^3-13x^2+30x
 INT = x^4/4 - 13x^3/3 + 15x^2 + C
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 y = 0 is the x-axis
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 There are 2 zeroes.
 x = 3 and x = 10
 The area from x = 0 to x = 3 is above the x-axis.
 x^4/4 - 13x^3/3 + 15x^2 + C
 f(0) = 0
 f(3) = 81/4 - 117 + 135
 f(3) - f(0) = 153/4 sq units
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 The area from x=3 to x=10 is below the x-axis.
 f(10) = 2500 - 13000/3 + 1500 = -1000/3
 f(10) - f(3) = -4459/12 sq units
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 IDK how you want to combine the two areas.