5.6 Riemann Sum - Introduction to Calculus

Riemann Sum

Suppose we wanted to find the area underneath the graph of a straight line that lies above the x-axis, between $\displaystyle x = a$ and $\displaystyle x = b$ . Since we have formulas for finding the area of basic shapes, we can easily figure this out.

\displaystyle f(x)=x+1

But what about finding the area underneath the graph of a general curve $\displaystyle y = f(x)$ and above the x-axis between $\displaystyle x = a$ and $\displaystyle x = b$ ?

Bernhard Riemann's idea was to carve up the desired area into rectangles and use their areas to estimate the true area. He called this the Riemann Sum and it goes as follows:

Create a partition $\displaystyle P$ , dividing up the interval $\displaystyle [a, b]$ into n subintervals $\displaystyle I_1, I_2, \dots, I_n$ .
Choose an x-value (call it $\displaystyle x_k$ ) in each subinterval $\displaystyle I_k$ .
For each $\displaystyle x_k$ we choose, draw a rectangle with height $\displaystyle f(x_k)$ and a width spanning $\displaystyle I_k$ ( $\displaystyle \Delta x_k$ ).

\displaystyle f(x)=2\sin(x+1)+4

The area of rectangle $\displaystyle k$ is:

\displaystyle f(x_k) \cdot \Delta x_k

The total area of all the rectangles is:

\displaystyle f(x_1) \cdot \Delta x_1 + f(x_2) \cdot \Delta x_2 + \dots + f(x_n) \cdot \Delta x_n = \sum_{k=1}^{n} f(x_k) \cdot \Delta x_k

Use the Riemann Sum to estimate the area below $\displaystyle y = \sin({1 \over 2}x)$ and above the x-axis, between $\displaystyle x = 0$ and $\displaystyle x = 2\pi$ . Use a partition of 4 equal subintervals and right endpoints.

\displaystyle y=\sin(x/2)

So, what happens as we make the rectangles skinnier?

\displaystyle f(x)=2+\sin(x-2)+\tfrac12\sin(3x)+\tfrac18\sin(5x)+\tfrac1{28}\sin(7x)

\displaystyle f(x)=2+\sin(x-2)+\tfrac12\sin(3x)+\tfrac18\sin(5x)+\tfrac1{28}\sin(7x)

\displaystyle f(x)=2+\sin(x-2)+\tfrac12\sin(3x)+\tfrac18\sin(5x)+\tfrac1{28}\sin(7x)

As the width of the rectangles decreases, the accuracy of the estimate increases. So, what happens if we let the width of all the rectangles in the partition approach 0?

Notationally:

\displaystyle \lim\limits_{\substack{n \rightarrow \infty \\ \left\|p\right\| \rightarrow 0}} \sum_{k=1}^{n} f(x_k) \cdot \Delta x_k

We should get the exact area, that is, our estimate is no longer just an estimate.

Notationally, instead of limit, we write it as:

THEOREM

Definite Integral

\displaystyle \begin{aligned}\int_{x = a}^{b} f(x) \ dx\end{aligned}

Start with the original exercise’s four right-endpoint rectangles and increase their number. Change the sample position and compare the sum, exact area, and error. Each finite refinement need not reduce the error, but as the largest width approaches zero, these continuous functions’ sums converge to the exact integral.

Source curveSample pointNumber of rectangles n4

\displaystyle f(x)=\sin(x/2)

\displaystyle \Delta x=1.57080,\qquad S_{4}=3.792238

Exact area: 4.000000 · Absolute error: 0.207762

QUESTION

For a continuous curve, what makes Riemann sums converge to the definite integral?

Riemann Sum

\displaystyle f(x)=x+1

But what about finding the area underneath the graph of a general curve $\displaystyle y = f(x)$ and above the x-axis between $\displaystyle x = a$ and $\displaystyle x = b$ ?

Bernhard Riemann's idea was to carve up the desired area into rectangles and use their areas to estimate the true area. He called this the Riemann Sum and it goes as follows:

Create a partition $\displaystyle P$ , dividing up the interval $\displaystyle [a, b]$ into n subintervals $\displaystyle I_1, I_2, \dots, I_n$ .
Choose an x-value (call it $\displaystyle x_k$ ) in each subinterval $\displaystyle I_k$ .
For each $\displaystyle x_k$ we choose, draw a rectangle with height $\displaystyle f(x_k)$ and a width spanning $\displaystyle I_k$ ( $\displaystyle \Delta x_k$ ).

\displaystyle f(x)=2\sin(x+1)+4

The area of rectangle $\displaystyle k$ is:

\displaystyle f(x_k) \cdot \Delta x_k

The total area of all the rectangles is:

\displaystyle f(x_1) \cdot \Delta x_1 + f(x_2) \cdot \Delta x_2 + \dots + f(x_n) \cdot \Delta x_n = \sum_{k=1}^{n} f(x_k) \cdot \Delta x_k

\displaystyle y=\sin(x/2)

So, what happens as we make the rectangles skinnier?

\displaystyle f(x)=2+\sin(x-2)+\tfrac12\sin(3x)+\tfrac18\sin(5x)+\tfrac1{28}\sin(7x)

\displaystyle f(x)=2+\sin(x-2)+\tfrac12\sin(3x)+\tfrac18\sin(5x)+\tfrac1{28}\sin(7x)

\displaystyle f(x)=2+\sin(x-2)+\tfrac12\sin(3x)+\tfrac18\sin(5x)+\tfrac1{28}\sin(7x)

As the width of the rectangles decreases, the accuracy of the estimate increases. So, what happens if we let the width of all the rectangles in the partition approach 0?

Notationally:

\displaystyle \lim\limits_{\substack{n \rightarrow \infty \\ \left\|p\right\| \rightarrow 0}} \sum_{k=1}^{n} f(x_k) \cdot \Delta x_k

We should get the exact area, that is, our estimate is no longer just an estimate.

Notationally, instead of limit, we write it as:

THEOREM

Definite Integral

\displaystyle \begin{aligned}\int_{x = a}^{b} f(x) \ dx\end{aligned}

Source curveSample pointNumber of rectangles n4

\displaystyle f(x)=\sin(x/2)

\displaystyle \Delta x=1.57080,\qquad S_{4}=3.792238

Exact area: 4.000000 · Absolute error: 0.207762

QUESTION

For a continuous curve, what makes Riemann sums converge to the definite integral?

Riemann Sum

Riemann Sum

Estimate the sine area

Definite Integral

Make rectangles approach the area under a curve

KNOWLEDGE QUIZ

Riemann Sum

Riemann Sum

Estimate the sine area

Definite Integral

Make rectangles approach the area under a curve

KNOWLEDGE QUIZ