Use the upper and lower Riemann sums to approximate the area of the region using `m=5` equal subintervals Estimate the area under the curve `y=v(1-x^2)`, `0<=x<=1`, `v` constant...
Use the upper and lower Riemann sums to approximate the area of the region using `m=5` equal subintervals
Estimate the area under the curve `y=v(1-x^2)`, `0<=x<=1`, `v` constant
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The upper sum `U` is the sum of the highest point of the function in each of the `m=5` intervals multiplied by the width of the intervals
`U = 0.2(f(0)+f(0.2)+f(0.4)+f(0.6)+f(0.8)) = `
`0.2v(1 + 0.96 + 0.84 + 0.64 + 0.36) = 0.760v`
The lower sum`L` is the sum of the lowest point of the function in each of the `m` intervals multiplied by the width of the intervals
`L = 0.2(f(0.2)+f(0.4)+f(0.6)+f(0.8)) = `
`0.2v(0.96+0.84+0.64+0.36+0) = 0.560v`
The integral is approximated by
`L+(U-L)/2 = 0.660v`
`int_0^1 v(1-x^2)dx approx 0.660v` answer
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