Calculating the area under a straight line can be done with geometry. Calculating the area under a curved line requires calculus. Often the area under a curve can be interpreted as the accumulated amount of whatever the function is modeling. Suppose a car’s speed in meters per second can be modeled by a quadratic for the first 8 seconds of acceleration:
\begin{align*}s(t)=t^2\end{align*}
How far has the car traveled in 8 seconds?
Watch This
http://www.youtube.com/watch?v=Z_OHgubPJKA James Sousa: Interpret the Meaning of Area Under a Function
http://www.youtube.com/watch?v=BvwyTXeuLS0 James Sousa: Approximate the Area Under a Curve with 4 Right Sided Rectangles
Guidance
The area under a curve can be approximated with rectangles equally spaced under a curve as shown below. For consistency, you can choose whether the boxes should hit the curve on the left hand corner, the right hand corner, the maximum value, or the minimum value. The more boxes you use the narrower the boxes will be and thus, the more accurate your approximation of the area will be.
The blue approximation uses right handed boxes. The red approximation assigns the height of the box to be the minimum value of the function in each subinterval. The green approximation assigns the height of the box to be the maximum value of the function in each subinterval. The yellow approximation uses left handed boxes. Rectangles above the \begin{align*}x\end{align*}-axis will have positive area and rectangles below the \begin{align*}x\end{align*}-axis will have negative area in this context.
All four of these area approximations get better as the number of boxes increase. In fact, the limit of each approximation as the number of boxes increases to infinity is the precise area under the curve.
This is where the calculus idea of an integral comes in. An integral is the limit of a sum as the number of summands increases to infinity.
\begin{align*}\int f(x)=\lim \limits_{n \to \infty}\sum \limits_{i=1}^n (Area \ of \ box \ i)\end{align*}
The symbol on the left is the calculus symbol of an integral. Using boxes to estimate the area under a curve is called a Riemann Sum.
Example A
Use four right handed boxes to approximate the area between 1 and 9 of the function \begin{align*}f(x)=\frac{1}{2}x-2\end{align*}.
Solution:
The area of the first box is 2 times the height of the function evaluated at 3:
\begin{align*}2 \cdot \left(\frac{1}{2} \cdot 3-2 \right)=3-4=-1\end{align*}
Because this box is under the \begin{align*}x\end{align*}-axis, its area is negative.
The area for each of the rest of the boxes is 2 times the height of the function evaluated at 5, 7 and 9.
\begin{align*}2 \cdot \left(\frac{1}{2} \cdot 5-2 \right)=5-4=1\end{align*}
\begin{align*}2 \cdot \left(\frac{1}{2} \cdot 7-2 \right)=7-4=3\end{align*}
\begin{align*}2 \cdot \left(\frac{1}{2} \cdot 9-2 \right)=9-4=5 \end{align*}
The approximate sum of the total area under the curve is: \begin{align*}-1+1+3+5=8\end{align*} square units.
Example B
Evaluate the exact area under the curve in Example A using the area formula for a triangle.
Solution: Remember that the area below the \begin{align*}x\end{align*} axis is negative while the area above the \begin{align*}x\end{align*} axis is positive.
Negative Area: \begin{align*}\frac{1}{2} \cdot 3 \cdot 1.5=\frac{9}{4}\end{align*}
Positive Area: \begin{align*}\frac{1}{2} \cdot 5 \cdot 2.5=\frac{25}{4}\end{align*}
Area under the curve between 1 and 8: \begin{align*}\frac{25}{4}-\frac{9}{4}=\frac{16}{4}=4\end{align*}
It appears that approximations that are 2 units wide produce an area with significant error.
Example C
Logan travels by bike at 20 mph for 3 hours. Then she gets in a car and drives 60 mph for 2 hours. Sketch both the distance vs. time graph and the rate vs. time graph. Use an area under the curve argument to connect the two graphs.
Solution: Distance vs. Time:
Rate vs. Time:
The slope of the first graph is 20 from 0 to 3 and then 60 from 3 to 5. The second graph is a graph of the slopes from the first graph. If you calculate the area of the second graph at the key points 0, 1, 2, 3, 4 and 5 you will see that they align perfectly with the points on the first graph.
\begin{align*}x\end{align*} | Area under curve from 0 to \begin{align*}x\end{align*} |
0 | 0 |
1 | 20 |
2 | 40 |
3 | 60 |
4 | 120 |
5 | 180 |
Concept Problem Revisited
You can use the area under the curve to find the total distance traveled in the first 8 seconds. Since the quadratic is a curve you must choose the number of subintervals you want to use and whether you want right or left handed boxes for estimating. Suppose you choose 8 left handed boxes of width one.
\begin{align*}x\end{align*} | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
Area of box to the right | 1⋅0 | 1⋅1 | 1⋅4 | 1⋅9 | 1⋅16 | 1⋅25 | 1⋅36 | 1⋅49 |
The approximate sum is \begin{align*}1+4+9+16+25+36+49=140\end{align*}. This means that the car traveled approximately 140 meters in the first 8 seconds.
Vocabulary
Subintervals are created when an interval is broken into smaller equally sized intervals.
An integral is the limit of a sum as the number of summands increases to infinity.
Using boxes to estimate the area under a curve is called a Riemann Sum.
A summand is one of many pieces being summed together.
Guided Practice
1. Approximate the area under the curve using eight subintervals and right endpoints.
\begin{align*}f(x)=3x^2-1, \ -1 \le x \le 7\end{align*}
2. Approximate the area under the curve using eight subintervals and left endpoints.
\begin{align*}f(x)=\frac{4}{x}+3, \ 2 \le x \le 6\end{align*}
3. Approximate the area under the curve using twenty subintervals and left endpoints.
\begin{align*}f(x)=x^x, \ 1 \le x \le 3\end{align*}
Answers:
1. While a graph is helpful to visualize the problem and drawing each box can help give meaning to each summand, it is not always necessary. Since there are going to be 8 subintervals over the total interval of \begin{align*}-1 \le x \le 7\end{align*}, each interval is going to have a width of 1. The height of each interval is going to be at the right hand endpoints of each subinterval (0, 1, 2, 3, 4, 5, 6, 7).
\begin{align*}\sum height \cdot width=\sum_{i=0}^7(3i^2-1) \cdot 1=412\end{align*}
2. Each interval will be only \begin{align*}\frac{1}{2}\end{align*} wide which means that the left endpoints have \begin{align*}x\end{align*} values of: 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5. Since the index of summation notation does not work with decimals, double each of these to get a good counting sequence: 4, 5, 6, 7, 8, 9, 10, 11 and remember to halve them in the argument of the summation.
\begin{align*}x&=\frac{i}{2}\\ \sum height \cdot width&=\sum \limits_{i=4}^{11} \left(\frac{4}{\left(\frac{i}{2}\right)} \right) \cdot \frac{1}{2}\approx 4.7462\end{align*}
3. When the number of subintervals gets large and the subintervals get extremely narrow it will be impossible to draw an accurate picture. This is why using summation notation and thinking through what the indices and the argument will be is incredibly important. With 20 subintervals between \begin{align*}[1,3]\end{align*}, each interval will be 0.1 wide. Left endpoints means that the first box has a height of \begin{align*}f(1)\end{align*} and the second box has a height of \begin{align*}f(1.1)\end{align*}.
\begin{align*}\sum height \cdot width &=f(1) \cdot 0.1+f(1.1) \cdot 0.1+f(1.2) \cdot 0.1+ \cdots +f(2.9) \cdot 0.1 \\ &=0.1(f(1)+f(1.1)+ \cdots f(2.9)) \\ &=0.1 \cdot \sum \limits_{i=10}^{29} f \left(\frac{i}{10} \right) \\ &=0.1 \cdot \sum \limits_{i=10}^{29} \left(\frac{i}{10} \right)^{\left(\frac{i}{10} \right)} \\ & \approx 12.47144\end{align*}
Your calculator can compute summations when you go under the math menu.