22. Riemann Sums, Integrals and the FTC

So far we have seen how to compute area using antiderivatives. However, mathematicians prefer to define the area under a curve by summing up smaller pieces of area. These are called Riemann sums.

a1. Right Riemann Sums and Areas

To find the area, \(A(a,b)\), under the graph of \(y=f(x)\) above the \(x\)-axis between \(x=a\) and \(x=b\), follow these steps:

def_right_riem_sum1
In these plots, \(a=2\) and \(b=14\).

Divide the interval \([a,b]\) into \(n\) subintervals of equal width: \[ \Delta x=\dfrac{b-a}{n} \] Then the partition points are: \[ x_i=a+i\Delta x \qquad \text{for} \quad i=0,1,\cdots,n \] In particular: \[ x_0=a \qquad \text{and} \qquad x_n=b \] and for the \(i^\text{th}\) subinterval, the left endpoint is \(x_{i-1}\) and the right endpoint is \(x_i\).

def_right_riem_sum2
Here, there are \(n=6\) partitions of width
\(\Delta x=\dfrac{14-2}{6}=2\).
And the partition points are
\(2, 4, 6, 8, 10, 12, 14\).

Evaluate the function at the right endpoint of each subinterval and multiply by the width of the subinterval to get: \[ f(x_i)\Delta x \] This is the area of the \(i^\text{th}\) rectangle which is an approximation to the area under the curve above the \(i^\text{th}\) subinterval. Add these up for all the subintervals to get the Riemann sum: \[ \sum_{i=1}^n f(x_i)\Delta x \] which is an approximation to the area under the curve above \([a,b]\). The approximation gets better as \(n\) gets larger.

def_right_riem_sum3
In this Riemann Sum, all \(6\) subintervals have
width \(\Delta x=2\) and the evaluation point
is the right endpoint of each interval.

Take the limit as the number of intervals, \(n\), goes to infinity. This is the definition of the area under the curve: \[ A(a,b) =\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x \] The limit of a Riemann sum on the right is an example of an integral.

Here, the number of rectangles
starts at \(6\) and keeps doubling.

The (definite) integral of a function \(f(x)\) is the limit of the Riemann sums: \[ \int_a^b f(x)\,dx =\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x \] The integral sign, \(\displaystyle \int\), (which is an elongated S for sum) replaces the limit of a sum. The differential, \(dx\), (previously discussed in the chapter on Differentials & Linear Approximation) replaces the change \(\Delta x\).

Thus the area under a curve is given by an integral:

If \(f(x)\) is positive, then the area under the graph of \(y=f(x)\) above the \(x\)-axis between \(x=a\) and \(x=b\) is: \[ A(a,b)=\int_a^b f(x)\,dx \]

The definitions of a Riemann sum and of an integral are actually more general and allow for functions, \(f(x)\), which are not necessarily positive, for partitions which are not necessarily evenly spaced and for evaluation points which are not necessarily the right endpoints. The full definition is given on the next page. However, for most purposes, the definitions on this page are totally sufficient.

Approximate the area below \(y=x^3\) above the \(x\)-axis between \(x=1\) and \(x=3\) using a Riemann sum with \(4\) equal width intervals and right endpoints.

eg_x^3

With \(n=4\) intervals, the width is: \[ \Delta x=\dfrac{3-1}{4}=0.5 \] The right endpoints are: \[ x_1=1.5 \quad x_2=2 \quad x_3=2.5 \quad x_4=3 \] and the function values are: \[\begin{aligned} f(1.5)&=3.375 & f(2)&=8 \\ f(2.5)&=15.625 & f(3)&=27 \end{aligned}\] So the Riemann sum is \[\begin{aligned} R_4&=3.375\cdot0.5+8\cdot0.5 \\ &\quad+15.625\cdot0.5+27\cdot0.5 \\ &=27.0 \end{aligned}\]

eg_x^3_right

Here is an exercise on approximating an integral using right Riemann sums. An analogous exercise using Riemann sums with other evaluation points appears on the next page. The process of taking the limit is discussed on the page after next.

Approximate the integral \(\displaystyle \int_1^9(x^2+1)\,dx\) using:

ex_x^2+1

  1. \(4\) equal width subintervals and the right endpoints of each interval.

    For equal width intervals \(\Delta x=\dfrac{b-a}{n}\). So: \(\displaystyle \int_a^bf(x)\,dx =\lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x\)

    \(R_4=336\)

    With \(n=4\) intervals, the width is \(\Delta x=\dfrac{9-1}{4}=2\). For right endpoints, the evaluation points are: \[ x_1=3 \qquad x_2=5 \qquad x_3=7 \qquad x_4=9 \] and the function values are: \[ f(3)=10 \qquad f(5)=26 \qquad f(7)=50 \qquad f(9)=82 \] So the Riemann sum is \[ R_4=10\cdot2+26\cdot2+50\cdot2+82\cdot2=336 \]

    ex_x^2+1_right_4

  2. \(8\) equal width subintervals and the right endpoints of each interval.

    \(R_8=292\)

    With \(n=8\) intervals, the width is \(\Delta x=\dfrac{9-1}{8}=1\). For right endpoints, the evaluation points are: \[\begin{aligned} x_1=2 \qquad x_2=3 \qquad x_3=4 \qquad x_4=5 \\ x_5=6 \qquad x_6=7 \qquad x_7=8 \qquad x_8=9 \end{aligned}\] and the function values are: \[\begin{aligned} f(2)=5\;\, \qquad f(3)=10 \qquad f(4)=17 \qquad f(5)=26 \\ f(6)=37 \qquad f(7)=50 \qquad f(8)=65 \qquad f(9)=82 \end{aligned}\] So the Riemann sum is \[ R_8=5+10+17+26+37+50+65+82=292 \]

    ex_x^2+1_right_8

In practice, one does not compute integrals using the Riemann sum formula. Rather, one uses this formula to derive the integral that needs to be computed in various applications (such as area, arc length and total mass). The integral is then computed using antiderivatives as prescribed by the Fundamental Theorem of Calculus: \[ \int_a^bf(x)\,dx=F(b)-F(a) \] as discussed later in this chapter.

© MYMathApps

Supported in part by NSF Grant #1123255