# 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 http://www.webassign.net/larson/4_02-30.gif 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 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`