Riemann Sums Assignment

Note: There are no graphs on this assignment.

Question 1

[2 points] Approximate the area under the graph of $f(x)=3x^2-9x+5$ on the interval $[-5,5]$ using left and right Riemann sums with $n=25$ and $n=50$ subintervals.

[The actual area is $300$.]

Question 2

[2 points] The area under the graph of $f(x)=\ln(\sin(x))$ from $x=1$ to $x=2$ is approximately $-0.0455$.

To get an idea of how big $n$ must be to get a good approximation (say correct to four decimal places), find both the left and right Riemann sums with $n=100$, $n=500$, and $n=1000$.

Question 3

[2 points] The graph of $x^2+y^2=25$ is a circle of radius 5 centered at the origin. From geometry, we know its area is $\pi\cdot5^2\approx78.54$. We will approximate this area using Riemann sums.

Let $\displaystyle f(x)=\sqrt{25-x^2}$ (the top half of the circle). Approximate the area between $f$ and the x-axis from $x=-5$ to $x=5$ using left and right Riemann sums with $n=100$ subintervals.

Now multiply this area by 2 to get an approximation for the area of the whole circle. You should be pretty close to the correct area.

Question 4

[3 points] Use CoCalc's sum command to evaluate the following sums. [Remember to declare variables.]

Part a

$\displaystyle\sum_{i=1}^{50}\frac{1}{i^2}$

Part b

$\displaystyle\sum_{k=10}^{100}\frac{k^3-3k^2}{5}$

Part c

$\displaystyle\sum_{k=1}^{n}\left(\left(\frac{k}{n}\right)^2+\frac{k}{n}\right)\cdot\frac{1}{n}$

[Hint: Declare both $n$ and $k$ to be variables.]

Question 5

[1 point] Use the limit command to calculate the limit as $n\to\infty$ of your answer from Question 4, Part c.

[Hint: The answer should be $\frac{5}{6}$]

After you finish Question 5, note: The limit in Question 5 gives the area between the x-axis and the function $f(x)=x^2+x$ over the interval from $x=0$ to $x=1$, because the sum in Question 4, Part c, is the right Riemann sum with $n$ rectangles for this function.

In other words, $\displaystyle\int_0^1 x^2+x\, dx=\lim_{n\to\infty}\sum_{k=1}^{n}\left(\left(\frac{k}{n}\right)^2+\frac{k}{n}\right)\cdot\frac{1}{n}=\frac{5}{6}$.