Riemann sums are used to approximate the area under a curve by dividing an interval into smaller subintervals and adding the areas of rectangles. The rectangle heights can come from the left endpoint, right endpoint, or midpoint of each subinterval. These approximations help build the idea of definite integrals and are especially useful when exact integration is difficult or when values are given in a table.
Notes
Left Riemann Sum
\(\displaystyle\int _a^b f(x)\,\)\(dx \approx \frac{b-a}{n} \cdot \left[f(a)+f\left(x_1\right)+f\left(x_2\right)+…+f\left(x_{n-1}\right)\right]\)
Right Riemann Sum
\(\displaystyle\int _a^b f(x)\,\)\(dx \approx \frac{b-a}{n} \cdot \left[f\left(x_1\right)+f\left(x_2\right)+…+f\left(x_{n-1}\right)+f(b)\right]\)
Midpoint Riemann Sum
\(\displaystyle\int _a^b f(x)\,\)\(dx \approx \frac{b-a}{n} \cdot \left[f\left(\frac{a+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+…+f\left(\frac{x_{n-1}+b}{2}\right)\right]\)
| \({\text{Overestimates and Underestimates}}\) |
| \(\) |
\(\underline{\text{Increasing Interval}}\) |
\(\underline{\text{Decreasing Interval}}\) |
| \(\underline{\text{Left Riemann}}\) |
\(\text{Underestimate}\) |
\(\text{Overestimate}\) |
| \(\underline{\text{Right Riemann}}\) |
\(\text{Overestimate}\) |
\(\text{Underestimate}\) |
Practice Problems
Solve each integral using Riemann sums with n subintervals.
\(\textbf{1)}\) Approximate using a Left Riemann sum with n=5
\(\displaystyle\int_{2}^{7}x^2 \,dx \,\,\, \)
The answer is \( 90 \)
\(\,\,\,\displaystyle\int _a^b f(x)\,dx \approx \frac{b-a}{n} \cdot \left[f(a)+f\left(x_1\right)+f\left(x_2\right)+…+f\left(x_{n-1}\right)\right]\)
\(\,\,\,\frac{7-2}{5}\left[f(2)+f(3)+f(4)+f(5)+f(6)\right]\)
\(\,\,\,\frac{5}{5}\left[f(2)+f(3)+f(4)+f(5)+f(6)\right]\)
\(\,\,\,1\left[f(2)+f(3)+f(4)+f(5)+f(6)\right]\)
\(\,\,\,\left[2^2+3^2+4^2+5^2+6^2\right]\)
\(\,\,\,\left[4+9+16+25+36\right]\)
\(\,\,\,\)The answer is \(90\)
\(\textbf{2)}\) Approximate using a Midpoint Riemann sum with n=3
\(\displaystyle\int_{0}^{6}x^3 \,dx \,\,\, \)
The answer is \( 306 \)
\(\,\,\,\displaystyle\int _a^b f(x)\,dx \approx \frac{b-a}{n} \cdot \left[f\left(\frac{a+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+…+f\left(\frac{x_{n-1}+b}{2}\right)\right]\)
\(\,\,\,\frac{6-0}{3} \cdot \left[f\left(\frac{0+2}{2}\right)+f\left(\frac{2+4}{2}\right)+f\left(\frac{4+6}{2}\right)\right]\)
\(\,\,\,\frac{6}{3} \cdot \left[f\left(\frac{2}{2}\right)+f\left(\frac{6}{2}\right)+f\left(\frac{10}{2}\right)\right]\)
\(\,\,\,2 \cdot \left[f\left(1\right)+f\left(3\right)+f\left(5\right)\right]\)
\(\,\,\,2 \cdot \left[1^3+3^3+5^3\right]\)
\(\,\,\,2 \cdot \left[1+27+125\right]\)
\(\,\,\,2 \cdot \left[153\right]\)
\(\,\,\,\)The answer is \(306\)
\(\textbf{3)}\) Approximate using a Right Riemann sum with n=4
\(\displaystyle\int_{0}^{2\pi}\sin{x} \,dx \,\,\, n=4 \)
The answer is \( 0 \)
\(\,\,\,\,\,\,\frac{b-a}{n} \cdot \left[f\left(x_1\right)+f\left(x_2\right)+…+f\left(x_{n-1}\right)+f(b)\right]\)
\(\,\,\,\,\,\,\frac{2\pi-0}{4} \cdot \left[f\left(\frac{\pi}{2}\right)+f\left(\pi\right)+f\left(\frac{3\pi}{2}\right)+f\left(2\pi\right)\right]\)
\(\,\,\,\,\,\,\frac{2\pi}{4} \cdot \left[\sin\left(\frac{\pi}{2}\right)+\sin\left(\pi\right)+\sin\left(\frac{3\pi}{2}\right)+\sin\left(2\pi\right)\right]\)
\(\,\,\,\,\,\,\frac{\pi}{2} \cdot \left[1+0+(-1)+0\right]\)
\(\,\,\,\,\,\,\frac{\pi}{2} \cdot \left[0\right]\)
\(\,\,\,\,\,\)The answer is \(0\)
\(\textbf{4)}\) Approximate using a Left Riemann sum with n=4
\(\displaystyle\int_{0}^{2}5-x^2 \,dx \,\,\, n=4 \)
The answer is \( 8.25 \)
\(\,\,\,\frac{b-a}{n} \cdot \left[f(a)+f(x_1)+f(x_2)+f(x_3)\right]\)
\(\,\,\,\frac{2-0}{4} \cdot \left[f(0)+f(0.5)+f(1)+f(1.5)\right]\)
\(\,\,\,\frac{2}{4} \cdot \left[5-0^2+5-(0.5)^2+5-1^2+5-(1.5)^2\right]\)
\(\,\,\,0.5 \cdot \left[5+4.75+4+2.75\right]\)
\(\,\,\,0.5 \cdot 16.5\)
\(\,\,\,\)The answer is \(8.25\)
\(\textbf{5)}\) Approximate using a Right Riemann sum with n=4
\(\displaystyle\int_{1}^{5}\displaystyle \frac{1}{x+2} \,dx \,\,\, n=4 \)
The answer is \( \approx 0.75952 \)
\(\,\,\,\Delta x=\frac{5-1}{4}=1\)
\(\,\,\,x_1=2,\;x_2=3,\;x_3=4,\;x_4=5\)
\(\,\,\,\displaystyle R_4=1\left[f(2)+f(3)+f(4)+f(5)\right]\)
\(\,\,\,\displaystyle R_4=1\left[\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right]\)
\(\,\,\,\displaystyle R_4\approx0.75952\)
\(\,\,\,\)The answer is \(\approx0.75952\)
\(\textbf{6)}\) Approximate using a Midpoint Riemann sum with n=4
\(\displaystyle\int_{1}^{3}x \sin{x} \,dx \,\,\, n=4 \)
The answer is \( \approx 2.85421 \)
\(\,\,\,\Delta x=\frac{3-1}{4}=0.5\)
\(\,\,\,\text{Midpoints: }1.25,\;1.75,\;2.25,\;2.75\)
\(\,\,\,\displaystyle M_4=0.5\left[f(1.25)+f(1.75)+f(2.25)+f(2.75)\right]\)
\(\,\,\,\displaystyle M_4=0.5\left[1.25\sin(1.25)+1.75\sin(1.75)+2.25\sin(2.25)+2.75\sin(2.75)\right]\)
\(\,\,\,\displaystyle M_4\approx2.85421\)
\(\,\,\,\)The answer is \(\approx2.85421\)
\(\textbf{7)}\) Approximate using a Left Riemann sum with n=4
\(\displaystyle\int_{0}^{6}\sqrt{x^2+4} \,dx \,\,\, n=4 \)
The answer is \( \approx 19.54497 \)
\(\,\,\,\Delta x=\frac{6-0}{4}=1.5\)
\(\,\,\,x_0=0,\;x_1=1.5,\;x_2=3,\;x_3=4.5\)
\(\,\,\,\displaystyle L_4=1.5\left[f(0)+f(1.5)+f(3)+f(4.5)\right]\)
\(\,\,\,\displaystyle L_4=1.5\left[\sqrt{0^2+4}+\sqrt{1.5^2+4}+\sqrt{3^2+4}+\sqrt{4.5^2+4}\right]\)
\(\,\,\,\displaystyle L_4=1.5\left[2+\sqrt{6.25}+\sqrt{13}+\sqrt{24.25}\right]\)
\(\,\,\,\)The answer is \(\approx19.54497\)
\(\textbf{8)}\) Approximate using a Right Riemann sum with n=4
\(\displaystyle\int_{0}^{4}\left(x^2+1\right) \,dx\)
The answer is \(34\)
\(\,\,\,\Delta x=\frac{4-0}{4}=1\)
\(\,\,\,x_1=1,\;x_2=2,\;x_3=3,\;x_4=4\)
\(\,\,\,\displaystyle R_4=1\left[f(1)+f(2)+f(3)+f(4)\right]\)
\(\,\,\,\displaystyle R_4=1\left[(2)+(5)+(10)+(17)\right]\)
\(\,\,\,\)The answer is \(34\)
\(\textbf{9)}\) Approximate using a Left Riemann sum with n=4
\(\displaystyle\int_{0}^{4}\left(x^2+1\right) \,dx\)
The answer is \(14\)
\(\,\,\,\Delta x=\frac{4-0}{4}=1\)
\(\,\,\,x_0=0,\;x_1=1,\;x_2=2,\;x_3=3\)
\(\,\,\,\displaystyle L_4=1\left[f(0)+f(1)+f(2)+f(3)\right]\)
\(\,\,\,\displaystyle L_4=1\left[(1)+(2)+(5)+(10)\right]\)
\(\,\,\,\)The answer is \(18\)
\(\textbf{10)}\) Approximate using a Midpoint Riemann sum with n=4
\(\displaystyle\int_{0}^{4}\left(x^2+1\right) \,dx\)
The answer is \(25\)
\(\,\,\,\Delta x=\frac{4-0}{4}=1\)
\(\,\,\,\text{Midpoints: }0.5,\;1.5,\;2.5,\;3.5\)
\(\,\,\,\displaystyle M_4=1\left[f(0.5)+f(1.5)+f(2.5)+f(3.5)\right]\)
\(\,\,\,\displaystyle M_4=1\left[1.25+3.25+7.25+13.25\right]\)
\(\,\,\,\)The answer is \(25\)
\(\textbf{11)}\) Approximate using a Left Riemann sum with n=6
\(\displaystyle\int_{0}^{3}\left(2x+1\right) \,dx\)
The answer is \(10.5\)
\(\,\,\,\Delta x=\frac{3-0}{6}=0.5\)
\(\,\,\,x_0=0,\;x_1=0.5,\;x_2=1,\;x_3=1.5,\;x_4=2,\;x_5=2.5\)
\(\,\,\,\displaystyle L_6=0.5\left[f(0)+f(0.5)+f(1)+f(1.5)+f(2)+f(2.5)\right]\)
\(\,\,\,\displaystyle L_6=0.5\left[1+2+3+4+5+6\right]\)
\(\,\,\,\)The answer is \(10.5\)
\(\textbf{12)}\) Approximate using a Right Riemann sum with n=6
\(\displaystyle\int_{0}^{3}\left(2x+1\right) \,dx\)
The answer is \(13.5\)
\(\,\,\,\Delta x=\frac{3-0}{6}=0.5\)
\(\,\,\,x_1=0.5,\;x_2=1,\;x_3=1.5,\;x_4=2,\;x_5=2.5,\;x_6=3\)
\(\,\,\,\displaystyle R_6=0.5\left[f(0.5)+f(1)+f(1.5)+f(2)+f(2.5)+f(3)\right]\)
\(\,\,\,\displaystyle R_6=0.5\left[2+3+4+5+6+7\right]\)
\(\,\,\,\)The answer is \(13.5\)
\(\textbf{13)}\) Approximate using a Midpoint Riemann sum with n=3
\(\displaystyle\int_{0}^{6}\sqrt{x+1} \,dx\)
The answer is \(\approx12.2515\)
\(\,\,\,\Delta x=\frac{6-0}{3}=2\)
\(\,\,\,\text{Midpoints: }1,\;3,\;5\)
\(\,\,\,\displaystyle M_3=2\left[f(1)+f(3)+f(5)\right]\)
\(\,\,\,\displaystyle M_3=2\left[\sqrt{2}+\sqrt{4}+\sqrt{6}\right]\)
\(\,\,\,\)The answer is \(\approx12.2515\)
\(\textbf{14)}\) Approximate using a Right Riemann sum with n=4
\(\displaystyle\int_{1}^{3}\frac{1}{x} \,dx\)
The answer is \(\approx0.95\)
\(\,\,\,\Delta x=\frac{3-1}{4}=0.5\)
\(\,\,\,x_1=1.5,\;x_2=2,\;x_3=2.5,\;x_4=3\)
\(\,\,\,\displaystyle R_4=0.5\left[\frac{1}{1.5}+\frac{1}{2}+\frac{1}{2.5}+\frac{1}{3}\right]\)
\(\,\,\,\displaystyle R_4=0.5\left[\frac{2}{3}+\frac{1}{2}+\frac{2}{5}+\frac{1}{3}\right]\)
\(\,\,\,\)The answer is \(\approx0.95\)
\(\textbf{15)}\) Approximate using a Left Riemann sum with n=4
\(\displaystyle\int_{1}^{3}\frac{1}{x} \,dx\)
The answer is \(\approx1.2833\)
\(\,\,\,\Delta x=\frac{3-1}{4}=0.5\)
\(\,\,\,x_0=1,\;x_1=1.5,\;x_2=2,\;x_3=2.5\)
\(\,\,\,\displaystyle L_4=0.5\left[\frac{1}{1}+\frac{1}{1.5}+\frac{1}{2}+\frac{1}{2.5}\right]\)
\(\,\,\,\displaystyle L_4=0.5\left[1+\frac{2}{3}+\frac{1}{2}+\frac{2}{5}\right]\)
\(\,\,\,\)The answer is \(\approx1.2833\)
\(\textbf{16)}\) Use a Left Riemann sum with the table values below to approximate \(\displaystyle\int_0^4 f(x)\,dx\).
\(\begin{array}{c|ccccc}x&0&1&2&3&4\\ \hline f(x)&3&5&4&6&8\end{array}\)
The answer is \(18\)
\(\,\,\,\Delta x=1\)
\(\,\,\,\displaystyle L_4=1\left[f(0)+f(1)+f(2)+f(3)\right]\)
\(\,\,\,\displaystyle L_4=1\left[3+5+4+6\right]\)
\(\,\,\,\)The answer is \(18\)
\(\textbf{17)}\) Use a Right Riemann sum with the table values below to approximate \(\displaystyle\int_0^4 f(x)\,dx\).
\(\begin{array}{c|ccccc}x&0&1&2&3&4\\ \hline f(x)&3&5&4&6&8\end{array}\)
The answer is \(23\)
\(\,\,\,\Delta x=1\)
\(\,\,\,\displaystyle R_4=1\left[f(1)+f(2)+f(3)+f(4)\right]\)
\(\,\,\,\displaystyle R_4=1\left[5+4+6+8\right]\)
\(\,\,\,\)The answer is \(23\)
\(\textbf{18)}\) Use a Midpoint Riemann sum with the table values below to approximate \(\displaystyle\int_0^8 f(x)\,dx\).
\(\begin{array}{c|ccccc}x&0&2&4&6&8\\ \hline f(x)&4&7&9&6&5\end{array}\)
The answer is \(30\)
\(\,\,\,\Delta x=4\)
\(\,\,\,\text{For }n=2\text{ on }[0,8]\text{, the midpoints are }2\text{ and }6.\)
\(\,\,\,\displaystyle M_2=4\left[f(2)+f(6)\right]\)
\(\,\,\,\displaystyle M_2=4\left[7+6\right]\)
\(\,\,\,\)The answer is \(52\)
\(\textbf{19)}\) Is the Left Riemann sum an overestimate or underestimate for \(f(x)=x^2+1\) on \([0,4]\)?
The answer is \(\text{underestimate}\)
\(\,\,\,f(x)=x^2+1\text{ is increasing on }[0,4].\)
\(\,\,\,\text{For an increasing function, a Left Riemann sum uses rectangle heights from the lower side of each subinterval.}\)
\(\,\,\,\)The answer is \(\text{underestimate}\)
\(\textbf{20)}\) Is the Right Riemann sum an overestimate or underestimate for \(f(x)=\frac{1}{x}\) on \([1,5]\)?
The answer is \(\text{underestimate}\)
\(\,\,\,f(x)=\frac{1}{x}\text{ is decreasing on }[1,5].\)
\(\,\,\,\text{For a decreasing function, a Right Riemann sum uses rectangle heights from the lower side of each subinterval.}\)
\(\,\,\,\)The answer is \(\text{underestimate}\)
See Related Pages\(\)
