009B Sample Midterm 1, Problem 5

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.

Foundations:
See the page on Riemann Sums.

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
$1(f(0)+f(1)+f(2))=1+0+-3=-2$.

(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
$1(f(1)+f(2)+f(3))=0+-3+-8=-11$.

(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
$\lim _{n\to \infty }{\frac {3}{n}}\sum _{i=1}^{n}f{\bigg (}i{\frac {3}{n}}{\bigg )}\,=\,\lim _{n\to \infty }{\frac {3}{n}}\sum _{i=1}^{n}{\bigg (}1-{\bigg (}i{\frac {3}{n}}{\bigg )}^{2}{\bigg )}$ .

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

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