document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "​f(x)=36-x^2​;​[​-1,​12]\r
\n" ); document.write( "\n" ); document.write( "The area between the​ x-axis and​ f(x) is \r
\n" ); document.write( "\n" ); document.write( "Please answer the area between the​ x-axis and​ f(x) \n" ); document.write( "
Algebra.Com's Answer #825563 by Alan3354(69443)\"\" \"About 
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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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\n" ); document.write( "​f(x)=36-x^2​;​[​-1,​12]
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\n" ); document.write( "x^2 = 36 ---> x-intercepts at -6 and +6
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\n" ); document.write( "Int(36 - x^2) = g(x) = 36x - x^3/3 + C
\n" ); document.write( "Area from -1 to +6 = g(6) - g(-1)
\n" ); document.write( "Area = 216 - (-36 + 1/3) = 179 2/3 sq units
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\n" ); document.write( "Area from 6 to 12:
\n" ); document.write( "Area = g(12) - g(6) = 288 sq units ------> you can do the calculations.
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\n" ); document.write( "The area from 6 to 12 is below the x-axis, sometimes viewed as negative.
\n" ); document.write( "Subtracting would give a negative area, so add them. \n" ); document.write( "
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