# 007B Sample Midterm 1, Problem 1 Detailed Solution

Let  $f(x)=1-x^{2}$ .

(a) Compute the left-hand Riemann sum approximation of  $\int _{0}^{3}f(x)~dx$ with  $n=3$ boxes.

(b) Compute the right-hand Riemann sum approximation of  $\int _{0}^{3}f(x)~dx$ with  $n=3$ boxes.

(c) Express  $\int _{0}^{3}f(x)~dx$ as a limit of right-hand Riemann sums (as in the definition of the definite integral). Do not evaluate the limit.

Background Information:
1. The height of each rectangle in the left-hand Riemann sum is given by choosing the left endpoint of the interval.
2. The height of each rectangle in the right-hand Riemann sum is given by choosing the right endpoint of the interval.

Solution:

(a)

Step 1:
Since our interval is  $[0,3]$ and we are using 3 rectangles, each rectangle has width 1.
So, the left-hand Riemann sum is
$1(f(0)+f(1)+f(2)).$ Step 2:
Thus, the left-hand Riemann sum is

${\begin{array}{rcl}\displaystyle {1(f(0)+f(1)+f(2))}&=&\displaystyle {1+0+-3}\\&&\\&=&\displaystyle {-2.}\end{array}}$ (b)

Step 1:
Since our interval is  $[0,3]$ and we are using 3 rectangles, each rectangle has width 1.
So, the right-hand Riemann sum is
$1(f(1)+f(2)+f(3)).$ Step 2:
Thus, the right-hand Riemann sum is

${\begin{array}{rcl}\displaystyle {1(f(1)+f(2)+f(3))}&=&\displaystyle {0+-3+-8}\\&&\\&=&\displaystyle {-11.}\end{array}}$ (c)

Step 1:
Let  $n$ be the number of rectangles used in the right-hand Riemann sum for  $f(x)=1-x^{2}.$ The width of each rectangle is
$\Delta x={\frac {3-0}{n}}={\frac {3}{n}}.$ Step 2:
So, the right-hand Riemann sum is
$\Delta x{\bigg (}f{\bigg (}1\cdot {\frac {3}{n}}{\bigg )}+f{\bigg (}2\cdot {\frac {3}{n}}{\bigg )}+f{\bigg (}3\cdot {\frac {3}{n}}{\bigg )}+\ldots +f(3){\bigg )}.$ Finally, we let  $n$ go to infinity to get a limit.
Thus,  $\int _{0}^{3}f(x)~dx$ is equal to  $\lim _{n\to \infty }{\frac {3}{n}}\sum _{i=1}^{n}f{\bigg (}i{\frac {3}{n}}{\bigg )}.$ (a)     $-2$ (b)     $-11$ (c)     $\lim _{n\to \infty }{\frac {3}{n}}\sum _{i=1}^{n}f{\bigg (}i{\frac {3}{n}}{\bigg )}$ 