# Calculus: Integrals I

## Right Riemann Sums

A right Riemann sum is just like a left Riemann sum, except the rectangles are anchored to the function by the top right corner. A picture is worth a thousand words:

You may be wonder if there is a pattern where left Riemann sums underestimate the area and right Riemann sums overestimate it, as seen above. You can quickly disabuse yourself of that idea by integrating a curve that slopes the other way:

The mathematical formulation of the right Riemann sum is almost identical to that of the left Riemann sum. However, the summation index start and stop values have both been increased by one in order to anchor the rectangle by the top right corner, rather than the top left corner:

$$\displaystyle\int\limits_{a}^{b} f(x) \, dx \approx \sum\limits_{i=1}^{N} f(a+\Delta x \cdot i) \Delta x$$

## Problems

1. Estimate $\displaystyle\int\limits_{0}^{2} 4x + 5 \, dx$   where   $N=4$

Step 1: Calculate $\Delta x$

$\Delta x = \dfrac{b - a}{N} \\ \Delta x = \dfrac{2-0}{4} \\ \Delta x = \dfrac{1}{2} \\$

Step 2: Calculate the sum:

$\displaystyle\int\limits_{0}^{2} 4x + 5 \, dx \approx \displaystyle\sum\limits_{i=1}^{N} f(0+i\cdot\Delta x) \, \Delta x \\ \displaystyle\int\limits_{0}^{2} 4x + 5 \, dx \approx \Delta x \displaystyle\sum\limits_{i=1}^{4} f(0+i\cdot\Delta x) \\ \displaystyle\int\limits_{0}^{2} 4x + 5 \, dx \approx \dfrac{1}{2} \left( \left( 4\dfrac{1}{2} + 5 \right)+ (4(1) + 5) + \left(4\dfrac{3}{2} + 5\right) + (4(2) + 5) \right) \\ \displaystyle\int\limits_{0}^{2} 4x + 5 \, dx \approx \dfrac{1}{2} \left( 7 + 9 + 11 + 13 \right) \\ \displaystyle\int\limits_{0}^{2} 4x + 5 \, dx \approx 16$

2. Estimate $\displaystyle\int\limits_{-1}^{1} x^2 \, dx$   where   $N=5$

Step 1: Calculate $\Delta x$

$\Delta x = \dfrac{b - a}{N} \\ \Delta x = \dfrac{1-(-1)}{5} \\ \Delta x = \dfrac{2}{5} \\$

Step 2: Calculate the sum:

$\displaystyle\int\limits_{-1}^{1} x^2 \, dx \approx \displaystyle\sum\limits_{i=1}^{N} f(-1+i\cdot\Delta x) \, \Delta x \\ \displaystyle\int\limits_{-1}^{1} x^2 \, dx \approx \Delta x \displaystyle\sum\limits_{i=1}^{5} f(-1+i\cdot\Delta x) \\ \displaystyle\int\limits_{-1}^{1} x^2 \, dx \approx \dfrac{2}{5} \left(\left(\dfrac{-3}{5}\right)^2+ \left(\dfrac{-1}{5}\right)^2 + \left(\dfrac{1}{5}\right)^2 + \left(\dfrac{3}{5}\right)^2 + 1^2 \right) \\ \displaystyle\int\limits_{-1}^{1} x^2 \, dx \approx \dfrac{2}{5} \left( \dfrac{9}{25} + \dfrac{1}{25} + \dfrac{1}{25} + \dfrac{9}{25} + 1 \right) \\ \displaystyle\int\limits_{-1}^{1} x^2 \, dx \approx \dfrac{2}{5} \left( \dfrac{45}{25} \right) \\ \displaystyle\int\limits_{-1}^{1} x^2 \, dx \approx \dfrac{90}{125} \\$

3. Estimate $\displaystyle\int\limits_{0}^{\pi} \sin(x) \, dx$   where   $N=4$

Step 1: Calculate $\Delta x$

$\Delta x = \dfrac{b - a}{N} \\ \Delta x = \dfrac{\pi-0}{4} \\ \Delta x = \dfrac{\pi}{4} \\$

Step 2: Calculate sum:

$\displaystyle\int\limits_{0}^{\pi} \sin(x) \, dx \approx \displaystyle\sum\limits_{i=1}^{N} f(0+i\cdot\Delta x) \, \Delta x \\ \displaystyle\int\limits_{0}^{\pi} \sin(x) \, dx \approx \displaystyle\Delta x \sum\limits_{i=1}^{4} f(0+i\cdot\Delta x) \\ \displaystyle\int\limits_{0}^{\pi} \sin(x) \, dx \approx \dfrac{\pi}{4} \left(\sin\left(\dfrac{\pi}{4}\right) + \sin\left(\dfrac{\pi}{2}\right) + \sin\left(\dfrac{3\pi}{4}\right) + \sin(\pi)\right) \\ \displaystyle\int\limits_{0}^{\pi} \sin(x) \, dx \approx \dfrac{\pi}{4} \left(\dfrac{\sqrt{2}}{2} + 1 + \dfrac{\sqrt{2}}{2} + 0 \right) \\ \displaystyle\int\limits_{0}^{\pi} \sin(x) \, dx \approx \dfrac{\pi}{4} \left( 1 + \sqrt{2} \right) \\ \displaystyle\int\limits_{0}^{\pi} \sin(x) \, dx \approx \dfrac{\pi + \pi\sqrt{2}}{4} \\$

4. Estimate $\displaystyle\int\limits_{0}^{2\pi} \cos(2\theta) \, d\theta$   where   $N=8$

Step 1: Calculate $\Delta \theta$

$\Delta \theta = \dfrac{b - a}{N} \\ \Delta \theta = \dfrac{2\pi - 0}{8} \\ \Delta \theta = \dfrac{\pi}{4} \\$

Step 2: Calculate sum:

$\displaystyle\int\limits_{0}^{2\pi} \cos(2\theta) \, d\theta \approx \displaystyle\sum\limits_{i=1}^{N} f(0+i\cdot\Delta \theta) \, \Delta \theta \\ \displaystyle\int\limits_{0}^{2\pi} \cos(2\theta) \, d\theta \approx \Delta \theta \displaystyle\sum\limits_{i=1}^{8} f(0+i\cdot\Delta \theta) \\ \displaystyle\int\limits_{0}^{2\pi} \cos(2\theta) \, d\theta \approx \dfrac{\pi}{4} \left( \cos\left(2\dfrac{\pi}{4}\right) + \cos\left(2\dfrac{\pi}{2}\right) + \cos\left(2\dfrac{3\pi}{4}\right) + \\ \cos(2\pi) + \cos\left(2\dfrac{5\pi}{4}\right) + \cos\left(2\dfrac{3\pi}{2}\right) + \cos\left(2\dfrac{7\pi}{4}\right) + \cos(2\cdot2\pi) \right) \\ \displaystyle\int\limits_{0}^{2\pi} \cos(2\theta) \, d\theta \approx \dfrac{\pi}{4} \left( \cos(0) + \cos\left(\dfrac{\pi}{2}\right) + \cos\left(\pi\right) + \cos\left(\dfrac{3\pi}{2}\right) + \\ \cos(2\pi) + \cos\left(\dfrac{\pi}{2}\right) + \cos\left(\pi\right) + \cos\left(\dfrac{3\pi}{2}\right) + \cos(0) \right) \\ \displaystyle\int\limits_{0}^{2\pi} \cos(2\theta) \, d\theta \approx \dfrac{\pi}{4} ( 1 + 0 - 1 + 0 + 1 + 0 - 1 + 0 + 1) \\ \displaystyle\int\limits_{0}^{2\pi} \cos(2\theta) \, d\theta \approx \dfrac{\pi}{4} ( 0 ) \\ \displaystyle\int\limits_{0}^{2\pi} \cos(2\theta) \, d\theta \approx 0$

5. Estimate $\displaystyle\int\limits_{-2}^{2} x^3 + 2x\, dx$   where   $N=7$

Step 1: Calculate $\Delta x$

$\Delta x = \dfrac{b - a}{N} \\ \Delta x = \dfrac{2-(-2)}{7} \\ \Delta x = \dfrac{4}{7} \\$

Step 2: Calculate sum:

$\displaystyle\int\limits_{-2}^{2} x^3 + 2x \, dx \approx \displaystyle\sum\limits_{i=1}^{N} f(-2+i\cdot\Delta x) \, \Delta x \\ \displaystyle\int\limits_{-2}^{2} x^3 + 2x \, dx \approx \Delta x \displaystyle\sum\limits_{i=0}^{6} f(-2+i\cdot\Delta x) \\ \displaystyle\int\limits_{-2}^{2} x^3 + 2x \, dx \approx \dfrac{4}{7} \left( \left( \left(\dfrac{-10}{7}\right)^3 + 2\dfrac{-10}{7} \right) + \left( \left(\dfrac{-6}{7}\right)^3 + 2\dfrac{-6}{7} \right) + \\ \left( \left(\dfrac{-2}{7}\right)^3 + 2\dfrac{-2}{7} \right) + \left( \left(\dfrac{2}{7}\right)^3 + 2\dfrac{2}{7} \right) + \left( \left(\dfrac{6}{7}\right)^3 + 2\dfrac{6}{7} \right) + \\ \left( \left(\dfrac{10}{7}\right)^3 + 2\dfrac{10}{7} \right) + \left( 2^3 + 2(2) \right) \right) \\ \displaystyle\int\limits_{-2}^{2} x^3 + 2x \, dx \approx \dfrac{4}{7} \left( \left( \dfrac{-1000}{343} + \dfrac{-20}{7} \right) + \left( \dfrac{-216}{343} + \dfrac{-12}{7} \right) + \\ \left( \dfrac{-8}{343} + \dfrac{-4}{7} \right) + \left( \dfrac{8}{343} + \dfrac{4}{7} \right) + \left( \dfrac{216}{343} + \dfrac{12}{7} \right) + \\ \left( \dfrac{1000}{343} + \dfrac{20}{7} \right) + (8+4) \right) \\$

Rather than calculate this all out, observe that there's a great deal of symmetry here. Notice that all but the first term can be negated by another term:

$\displaystyle\int\limits_{-2}^{2} x^3 + 2x \, dx \approx \dfrac{4}{7} \left( - \left( \dfrac{1000}{343} + \dfrac{20}{7} \right) - \left( \dfrac{216}{343} + \dfrac{12}{7} \right) - \\ \left( \dfrac{8}{343} + \dfrac{4}{7} \right) + \left( \dfrac{8}{343} + \dfrac{4}{7} \right) + \left( \dfrac{216}{343} + \dfrac{12}{7} \right) + \\ \left( \dfrac{1000}{343} + \dfrac{20}{7} \right) + 12 \right) \\ \displaystyle\int\limits_{-2}^{2} x^3 + 2x \, dx \approx \dfrac{4}{7} ( 12 ) \\ \displaystyle\int\limits_{-2}^{2} x^3 + 2x \, dx \approx \dfrac{48}{7}$

6. Estimate $\displaystyle\int\limits_{1}^{2} \dfrac{2}{x} \, dx$   where   $N = 5$

A calculator may be of service for this problem.

Step 1: Calculate $\Delta x$

$\Delta x = \dfrac{b - a}{N} \\ \Delta x = \dfrac{2-1}{5} \\ \Delta x = \dfrac{1}{5} \\$

Step 2: Calculate sum:

$\displaystyle\int\limits_{1}^{2} \dfrac{2}{x} \, dx \approx \displaystyle\sum\limits_{i=1}^{N} f(1+i\cdot\Delta x) \, \Delta x \\ \displaystyle\int\limits_{1}^{2} \dfrac{2}{x} \, dx \approx \Delta x \displaystyle\sum\limits_{i=1}^{5} f(1+i\cdot\Delta x) \\ \displaystyle\int\limits_{1}^{2} \dfrac{2}{x} \, dx \approx \dfrac{1}{5} \left(\dfrac{2}{1+1/5} + \dfrac{2}{1+2/5} + \dfrac{2}{1+3/5} + \dfrac{2}{1+4/5} + \dfrac{2}{2} \right) \\ \displaystyle\int\limits_{1}^{2} \dfrac{2}{x} \, dx \approx\approx \dfrac{1}{5} \left( \dfrac{1627}{252} \right) \\ \displaystyle\int\limits_{1}^{2} \dfrac{2}{x} \, dx \approx\approx \dfrac{7127}{1260} \\$

7. Estimate $\displaystyle\int_{-4}^{-1} 3x^4 + 7x + 2 \, dx$   where   $N=6$

A calculator may be of service for this problem.

Step 1: Calculate $\Delta x$

$\Delta x = \dfrac{b - a}{N} \\ \Delta x = \dfrac{-1-(-4)}{6} \\ \Delta x = \dfrac{3}{6} \\ \Delta x = \dfrac{1}{2} \\$

Step 2: Calculate sum:

$\displaystyle\int\limits_{-4}^{-1} 3x^4 + 7x + 2 \, dx \approx \displaystyle\sum\limits_{i=1}^{N} f(-4+i\cdot\Delta x) \, \Delta x \\ \displaystyle\int\limits_{-4}^{-1} 3x^4 + 7x + 2 \, dx \approx \Delta x \displaystyle\sum\limits_{i=1}^{6} f(-4+i\cdot\Delta x) \\ \displaystyle\int\limits_{-4}^{-1} 3x^4 + 7x + 2 \, dx \approx \dfrac{1}{2} \left( \left( 3\left(\dfrac{-7}{2}\right)^4 + 7\left(\dfrac{-7}{2}\right) + 2 \right) + \\ \left( 3(-3)^4 + 7(-3) + 2 \right) + \left( 3\left(\dfrac{-5}{2}\right)^4 + 7\left(\dfrac{-5}{2}\right) + 2 \right) + \\ \left( 3(-2)^4 + 7(-2) + 2 \right) + \left( 3\left(\dfrac{-3}{2}\right)^4 + 7\left(\dfrac{-3}{2}\right) + 2 \right) + \left( 3(-1)^4 + 7(-1) + 2 \right) \right) \\ \displaystyle\int\limits_{-4}^{-1} 3x^4 + 7x + 2 \, dx \approx \dfrac{1}{2} \left( \left( \dfrac{7203}{16}-\dfrac{49}{2} + 2 \right) + \left( 243 - 21 + 2 \right) + \\ \left( \dfrac{1875}{16} - \dfrac{35}{2} + 2 \right) + \left( 48 - 14 + 2 \right) + \left( \dfrac{243}{16} - \dfrac{21}{2} + 2 \right) + \left( 3 - 7 + 2 \right) \right) \\ \displaystyle\int\limits_{-4}^{-1} 3x^4 + 7x + 2 \, dx \approx \dfrac{1}{2} \left( \dfrac{12705}{16} \right) \\ \displaystyle\int\limits_{-4}^{-1} 3x^4 + 7x + 2 \, dx \approx \dfrac{12705}{32} \\$

8. Estimate $\displaystyle\int\limits_{0}^{8} 2^{x} \, dx$   where   $N=4$

Step 1: Calculate $\Delta x$

$\Delta x = \dfrac{b - a}{N} \\ \Delta x = \dfrac{8 - 0}{4} \\ \Delta x = 2 \\$

Step 2: Calculate sum:

$\displaystyle\int\limits_{0}^{8} 2^x \, dx \approx \displaystyle\sum\limits_{i=1}^{N} f(0+i\cdot \Delta x) \, \Delta x \\ \displaystyle\int\limits_{0}^{8} 2^x \, dx \approx \Delta x \displaystyle\sum\limits_{i=1}^{4} f(0+i\cdot \Delta x) \\ \displaystyle\int\limits_{0}^{8} 2^x \, dx \approx 2 \left( 2^2 + 2^4 + 2^6 + 2^8 \right) \\ \displaystyle\int\limits_{0}^{8} 2^x \, dx \approx \left( 2^3 + 2^5 + 2^7 + 2^9 \right) \\ \displaystyle\int\limits_{0}^{8} 2^x \, dx \approx 680 \\$

9. Estimate $\displaystyle\int\limits_{0}^{8} 2^{x} \, dx$   where   $N=8$

Step 1: Calculate $\Delta x$

$\Delta x = \dfrac{b - a}{N} \\ \Delta x = \dfrac{8 - 0}{8} \\ \Delta x = 1 \\$

Step 2: Calculate sum:

$\displaystyle\int\limits_{0}^{8} 2^x \, dx \approx \displaystyle\sum\limits_{i=1}^{N} f(0 + i \cdot \Delta x) \, \Delta x \\ \displaystyle\int\limits_{0}^{8} 2^x \, dx \approx \Delta x \displaystyle\sum\limits_{i=1}^{9} f(0 + i \cdot \Delta x) \\ \displaystyle\int\limits_{0}^{8} 2^x \, dx \approx 1 \left( 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 + 2^8 \right) \\ \displaystyle\int\limits_{0}^{8} 2^x \, dx \approx 510 \\$

10. Estimate $\displaystyle\int\limits_{a}^{b} C \, dx$   where   $N=N_{0} \geq 1$.

Based on what you know about geometry, what is the exact solution? Does the left Riemann sum produce an exact result?

Step 1: Calculate $\Delta x$

$\Delta x = \dfrac{b - a}{N} \\ \Delta x = \dfrac{b - a}{N_{0}} \\$

Step 2: Calculate sum:

$\displaystyle\int\limits_{a}^{b} C \, dx \approx \displaystyle\sum\limits_{i=1}^{N} C \, \Delta x \\ \displaystyle\int\limits_{a}^{b} C \, dx \approx \Delta x C \displaystyle\sum\limits_{i=1}^{N_0} 1 \\ \displaystyle\int\limits_{a}^{b} C \, dx \approx \dfrac{b-a}{N_0} C \left( N_{0} \right) \\ \displaystyle\int\limits_{a}^{b} C \, dx \approx C(b - a) \\$

The graph is a rectangle, and geometry tells us that the area of a rectangle is $\text{width}*\text{height}$. Here, the height of the rectangle is $C$ and the width is $b-a$. We can see that the result of the Riemann sum is exact:

$\displaystyle\int\limits_{a}^{b} C \, dx = \displaystyle\sum\limits_{i=1}^{N_0} C \, \Delta x = C(b-a) \\$

This shouldn't be surprising. The left Riemann sum uses little rectangles to estimate the area under a curve. In this case, however, the curve being split up is itself a rectangle.