SOLUTION: Use a graphing utility, where helpful, to find the area of the region enclosed by the curves. y=x^3−9x^2+14x and y=0. Carry out all calculations exactly and round to 2

Question 1075206: Use a graphing utility, where helpful, to find the area of the region enclosed by the curves.
y=x^3−9x^2+14x and y=0.
Carry out all calculations exactly and round to 2 decimal places the final answer only.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
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.
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SOLUTION: Use a graphing utility, where helpful, to find the area of the region enclosed by the curves. y=x^3&#8722;9x^2+14x and y=0. Carry out all calculations exactly and round to 2

SOLUTION: Use a graphing utility, where helpful, to find the area of the region enclosed by the curves. y=x^3−9x^2+14x and y=0. Carry out all calculations exactly and round to 2