Question 124501
The line y=4 meets the y=sqrt(x) curve at (16,0).
Is the following what you have as the area under
the line,above the curve, and in the 1st Quadrant?
Area = integral from 0 to 16 of 4dx - integral from 0 to 16 of x^(1/2)dx?
= 64 - (2/3)(64)
= (1/3)64
= 21 1/3
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Cheers,
Stan H.