Question

Evaluate the upper and lower sums for the function f(x) = x² + 1 on the interval [0, 2] with n = 4. Illustrate with diagrams.

Answer 1

For the function
`\(f(x) = x^2 + 1\)` on the interval
`\([0, 2]\)` with (n = 4), the upper sum is
`\(U_4 = (17)/(4)\)` and the lower sum is
`\(L_4 = (11)/(4)\)`.

Step-by-step explanation:

To evaluate the upper and lower sums for the function
`\(f(x) = x^2 + 1\)` on the interval
`\([0, 2]\)` with (n = 4), we first divide the interval into (4) subintervals of equal width. The width of each subinterval, denoted as
`\(\Delta x\)`, is calculated as
`\((2 - 0)/(4) = (1)/(2)\)`.

For each subinterval
`\([x_(i-1), x_i]\)`, we find the upper sum by taking the supremum of
`\(f(x)\)` on that interval and multiplying it by
`\(\Delta x\)`. The lower sum is similarly obtained by taking the infimum of
`\(f(x)\)` on the interval and multiplying it by
`\(\Delta x\)`.

The upper sum
`\(U_4\) is calculated as \[U_4 = (1)/(2)\left((1 + 1) + \left((9)/(4) + 1\right) + \left(4 + 1\right) + \left((25)/(4) + 1\right)\right) = (17)/(4),\]` and the lower sum
`\(L_4\)` is calculated as
`\[L_4 = (1)/(2)\left((1 + 1) + \left(1 + 1\right) + \left(1 + 1\right) + \left(1 + 1\right)\right) = (11)/(4).\]`

These sums represent overestimation and underestimation, respectively, of the integral of
`\(f(x)\)` over the given interval. The diagrams illustrating the upper and lower sums visually demonstrate the areas of rectangles over and under the curve, providing a geometric understanding of the Riemann sum approximation.