# 009B Sample Midterm 1, Problem 1 Detailed Solution

Let  ${\displaystyle f(x)=1-x^{2}}$.

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

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

(c) Express  ${\displaystyle \int _{0}^{3}f(x)~dx}$  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.
3. See the Riemann sums (insert link) for more information.

Solution:

(a)

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

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

(b)

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

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

(c)

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

(a)     ${\displaystyle -2}$
(b)     ${\displaystyle -11}$
(c)     ${\displaystyle \lim _{n\to \infty }{\frac {3}{n}}\sum _{i=1}^{n}f{\bigg (}i{\frac {3}{n}}{\bigg )}}$