Asked by Anonymous
Consider the straight line: y=1x+3. Calculate the area UNDER this line which is bounded by y=0, x=8 and x=15. (That is, the area from the horizontal axis (y=0) to the line, between the values of x=8 and x=15.) Give you answer to the nearest whole number, rounding up
Answers
Answered by
Reiny
Using Calculus:
Area = ∫(x+3) dx from 8 to 15
= [x^2/2 + 3x] from 8 to 15
= 225/2 + 45 - 64/2 - 24
= 203/2 or 101.5
of course we don't need Calculus to do this, since the shape is a simple trapezoid
at x=8, height = 11
at x = 15 , height = 18
distance between = 15-8 = 7
Area = (1/2)(11+18)(7) = 101.5
Area = ∫(x+3) dx from 8 to 15
= [x^2/2 + 3x] from 8 to 15
= 225/2 + 45 - 64/2 - 24
= 203/2 or 101.5
of course we don't need Calculus to do this, since the shape is a simple trapezoid
at x=8, height = 11
at x = 15 , height = 18
distance between = 15-8 = 7
Area = (1/2)(11+18)(7) = 101.5
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