Question 1193521
Use the definite integral to find the area between the​ x-axis and​ f(x) over the indicated interval. Check first to see if the graph crosses the​ x-axis in the given interval.
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​f(x)=36-x^2​;​[​-1,​12]
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x^2 = 36 ---> x-intercepts at -6 and +6
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Int(36 - x^2) = g(x) = 36x - x^3/3 + C
Area from -1 to +6 = g(6) - g(-1)
Area = 216 - (-36 + 1/3) = 179 2/3 sq units
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Area from 6 to 12:
Area = g(12) - g(6) = 288 sq units ------> you can do the calculations.
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The area from 6 to 12 is below the x-axis, sometimes viewed as negative.
Subtracting would give a negative area, so add them.