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Wize University Calculus 2 Textbook > Integration

Riemann Sums

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Sigma Notation

Notation
The symbol ∑i=mnai\displaystyle \sum_{i=m}^na_i denotes the finite sum of ai′sa_i's, where the index ii starts at mm and ends at nn.
∑i=mnai=am+am+1+am+2+.....+an\displaystyle \sum_{i=m}^na_i=a_m+a_{m+1}+a_{m+2}+.....+a_n

Properties
  • ∑i=mnai+∑i=mnbi=∑i=mn(ai+bi)\displaystyle \sum_{i=m}^na_i+\sum_{i=m}^nb_i=\sum_{i=m}^n\left(a_{i+}b_i\right)
  • k ∑i=mnai=∑i=mn(kai)\displaystyle k\ \sum_{i=m}^na_i=\sum_{i=m}^n\left(ka_i\right)

Important Sums to Memorize!
∑i=1n 1n∑i=1n in(n+1)2∑i=1n i2n(n+1)(2n+1)6∑i=1n i3(n(n+1)2)2\begin{array}{|c|c|} \hline \displaystyle \sum_{i=1}^n\ 1&n\\\hline \displaystyle \sum_{i=1}^n\ i&\displaystyle \frac{n\left(n+1\right)}{2}\\\hline \displaystyle ^{ }\sum_{i=1}^n\ i^2&\displaystyle\frac{n\left(n+1\right)\left(2n+1\right)}{6}\\\hline \displaystyle \sum_{i=1}^n\ i^3&\displaystyle \left(\frac{n\left(n+1\right)}{2}\right)^2\\\hline \end{array}
*For these sums, the index must start at 1!

Wize Tip
If you have a sum that has a form that is very different than these important sums, try expanding out the first few terms to see what the sum looks like.

Sometimes, the "middle terms" will all cancel out -- this is called a telescoping sum.

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Example
Evaluate the sum ∑i=020cos⁡(π2i)\displaystyle \sum_{i=0}^{20}\cos\left(\frac{\pi}{2}i\right)
Since we don't have a formula for sums of this form, let's expand out a few terms:
=cos⁡(0)+cos⁡(π2)+cos⁡(π)+cos⁡(3π2)+cos⁡(2π)+...+cos⁡(10π)=\cos\left(0\right)+\cos\left(\frac{\pi}{2}\right)+\cos\left(\pi\right)+\cos\left(\frac{3\pi}{2}\right)+\cos\left(2\pi\right)+...+\cos\left(10\pi\right)

Notice that the sum of the first 4 terms is 1+0+(−1)+0=01+0+\left(-1\right)+0=0, and this pattern happens for the next 4 terms, and the next 4 terms, and so on.

So, we know that ∑i=019cos⁡(π2i)=0\displaystyle\sum_{i=0}^{19}\cos\left(\frac{\pi}{2}i\right)=0

Therefore, ∑i=020cos⁡(π2i)=cos⁡(10π)=1\displaystyle \sum_{i=0}^{20}\cos\left(\frac{\pi}{2}i\right)=\cos(10\pi)=1.

Practice: Sigma Notation

Evaluate ∑i=1020 (3i−4)\displaystyle \sum_{i=10}^{20}\ \left(3i-4\right).


Practice: Sigma w/ Limits

Evaluate lim⁡n→∞ ∑i=1n(3n4)⋅[i(2i2−1)+1]\displaystyle \lim_{n\rightarrow\infty}\ \sum_{i=1}^n\left(\frac{3}{n^4}\right)\cdot\left[i\left(2i^2-1\right)+1\right]

Practice: Telescoping Sums

Evaluate ∑i=12000 (1i+1−1i+2)\displaystyle \sum_{i=1}^{2000}\ \left(\frac{1}{i+1}-\frac{1}{i+2}\right)
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Riemann Sums

To estimate the area bounded by a curve and the x-axis on a certain interval, we can "slice" the area into thin rectangles and add up the areas of those rectangles--called Riemann Sum.

Left-Riemann Sum
∑i=1n f(a+(i−1)Δx)×Δx\displaystyle\sum_{i=1}^n\ f\left(a+\left(i-1\right)\Delta x\right)\times\Delta x, where Δx=b−an\displaystyle\Delta x=\frac{b-a}{n}

Right-Riemann Sum
∑i=1n f(a+iΔx)×Δx\displaystyle\sum_{i=1}^n\ f\left(a+i\Delta x\right)\times\Delta x, where Δx=b−an\displaystyle\Delta x=\frac{b-a}{n}

Exact Area Under a Curve
The area under the curve y=f(x)y=f\left(x\right), between x=ax=a and x=bx=b is given by
A=lim⁡n→∞∑i=1n f(xi∗)×Δx\displaystyle A=\lim_{n\rightarrow\infty}\sum_{i=1}^n\ f\left(x_i^{\ast}\right)\times\Delta x

Wize Tip
*Unless we are told otherwise, the question will typically use xi ∗=a+iΔxx_i^{\ \ast}=a+i\Delta x

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Example
Describe the area calculated by lim⁡n→∞ ∑i=1n[1n(1+in)2]\displaystyle \lim_{n\rightarrow\infty}\ \sum_{i=1}^n\left[\frac{1}{n}\left(1+\frac{i}{n}\right)^2\right] .

Matching this up with A=lim⁡n→∞∑i=1n f(xi∗)×Δx\displaystyle A=\lim_{n\rightarrow\infty}\sum_{i=1}^n\ f\left(x_i^{\ast}\right)\times\Delta x, we see that
  • iΔx=i(b−a)n=in → b−a=1i\Delta x=\frac{i\left(b-a\right)}{n}=\frac{i}{n}\ \to\ b-a=1
  • f(xi ∗)⋅Δx=f(a+iΔx)⋅Δx=f(a+in)⋅1n=1n⋅(1+in)2f\left(x_i^{\ \ast}\right)\cdot \Delta x=f\left(a+i\Delta x\right)\cdot \Delta x=f\left(a+\frac{i}{n}\right)\cdot\frac{1}{n}=\frac{1}{n}\cdot\left(1+\frac{i}{n}\right)^2 → a=1a=1 and f(x)=x2f\left(x\right)=x^2
  • Meaning that b−1=1 →b=2b-1=1\ \to b=2
Therefore, this limit describes the area under the curve f(x)=x2f\left(x\right)=x^2 on the interval [1, 2]\left[1,\ 2\right].

Practice: Estimating Area Using Riemann Sum

Given the following table of values for the function f(x)f\left(x\right), find the left-endpoint Riemann sum L3L_3 and right-endpoint Riemann sum R3R_3 to estimate the area under the graph of f(x)f\left(x\right) on the interval [0, 6]\left[0,\ 6\right].

Practice: Area using Riemann Sums

The area bounded by the function f(x)f\left(x\right) and the xx-axis in the interval [a, b]\left[a,\ b\right] is represented by lim⁡n→∞ ∑i=1n(3n)[cos⁡(2+3in)(2+3in)]\displaystyle \lim_{n\rightarrow\infty}\ \sum_{i=1}^n\left(\frac{3}{n}\right)\left[\frac{\cos\left(2+\frac{3i}{n}\right)}{\left(2+\frac{3i}{n}\right)}\right].
Find the function f(x)f\left(x\right) and the interval [a, b]\left[a,\ b\right].

The area bounded by the function f(x)f\left(x\right) and the xx-axis in the interval [a,b]\left[a,b\right] is given by lim⁡n→∞∑i=1n4in2[ein]\displaystyle \lim_{n\to\infty}\sum_{i=1}^n\frac{4i}{n^2}\left[e^{\frac{i}{n}}\right].

Determine f(x)f(x), aa, and bb.
Extra Practice
Practice: Definite Integral Using Riemann Sums
Evaluate ∫04(2x2+3)dx\displaystyle \int_{0}^{4}(2x^2 +3)\text{d}x using Riemann sums.
Riemann Sums
Use the left-endpoint Riemann sum, with n=4n=4, to estimate the area under the curve f(x)=ex−xf\left(x\right)=e^x-x on the interval [1, 13]\left[1,\ 13\right].
Riemann Sums: Approximating Areas
The area bounded by the function f(x)f\left(x\right) and the xx-axis in the interval [a,b]\left[a,b\right] is given by lim⁡n→∞∑i=1n4in2[ein]\displaystyle \lim_{n\to\infty}\sum_{i=1}^n\frac{4i}{n^2}\left[e^{\frac{i}{n}}\right].
Determine f(x)f(x), aa, and bb.
Practice: Definite Integral Using Riemann Sums
Q.\textbf{Q.} Evaluate ∫04(2x2+3)dx\displaystyle \int_{0}^{4}(2x^2 +3)\text{d}x using Riemann sums.