Approximate areas under graphs using trapezoidal rule
When a curve forms a familiar geometric shape, it's relatively easy to determine the area enclosed between the curve and the $x$x-axis.
But what if the curve you've graphed doesn't make a nice geometric shape? One approach is then to approximate the area between the curve and the $x$x-axis.
Under-approximating the area under a curve
Let's consider finding the area between the curve $y=x^2+1$y=x2+1 and the $x$x-axis from $x=0$x=0 to $x=1$x=1.
We're really stepping into the shoes of the Leibniz and Newton here as we put on our mathematician thinking caps!
How do you approximate a curved area? Using circles? But how?
Let's go back to primary school. How did you first experiment with the idea of the area of a shape? You counted the number of square centimetres inside the shape, often giving an approximation.
We'll take a similar approach, but instead of counting squares, we'll find the areas of rectangles.
In this first approach we'll under-approximate the area and to do so we'll draw a series of $5$5 rectangles that sit underneath the curve. Why $5$5 rectangles? No particular reason, it's just a nice place to start.
To calculate the total area we find the area of each rectangle and add them all up.
We can see the width of each is $0.2$0.2, but what is the height? To find the height, we need to substitute the $x$x value into the function to calculate the $y$y value and hence the height of each rectangle.
$Area=0.2\left(0^2+1\right)+0.2\left(0.2^2+1\right)+0.2\left(0.4^2+1\right)+0.2\left(0.6^2+1\right)+0.2\left(0.8^2+1\right)$Area=0.2(02+1)+0.2(0.22+1)+0.2(0.42+1)+0.2(0.62+1)+0.2(0.82+1)
$Area=0.2\left(1+1.04+1.16+1.36+1.64\right)$Area=0.2(1+1.04+1.16+1.36+1.64)
$Area=1.24$Area=1.24 $units^2$units2
If we wanted a better approximation and less white space between the rectangles and the curve, what could we do? We could increase the number or rectangles, making the width of each smaller.
Let's try this with$10$10rectangles.
$Area=0.1\left(1+1.01+1.04+1.09+1.16+1.25+1.36+1.49+1.64+1.81\right)$Area=0.1(1+1.01+1.04+1.09+1.16+1.25+1.36+1.49+1.64+1.81)
$Area=1.285$Area=1.285 $units^2$units2
We could do more and more rectangles to get an even better approximation. Experiment with the applet below to see what happens to the under-approximation of the area.
Over-approximating the area under a curve
We can also choose to over-approximate instead of under-approximate our area.
Again I've chosen $5$5 rectangles, and notice the placement of the right hand side of the rectangle so that the rectangles are drawn over the curve.
$Area=0.2\left(1.04+1.16+1.36+1.64+2\right)$Area=0.2(1.04+1.16+1.36+1.64+2)
$Area=1.4325$Area=1.4325 $units^2$units2
The middle way for approximating area
We can also use a middle-way here and draw rectangles so that the middle of the rectangle touches the curve, as shown above.
$Area=0.2\left(1.01+1.09+1.25+1.47+1.81\right)$Area=0.2(1.01+1.09+1.25+1.47+1.81)
$Area=1.33$Area=1.33 $units^2$units2
Worked Examples
Question 1
The function $f\left(x\right)=5x$f(x)=5x is defined on the interval $\left[0,6\right]$[0,6].
Graph $f\left(x\right)$f(x).
Loading Graph...Find the area $A$A under $f\left(x\right)$f(x) by partitioning $\left[0,6\right]$[0,6] into three sub-intervals of equal length and using rectangles for each sub-interval whose heights are equal to the function values of the left side of the sub-intervals.
Find the area $A$A under $f\left(x\right)$f(x) by partitioning $\left[0,6\right]$[0,6] into three subintervals of equal length and using rectangles whose widths are equal to the range of the intervals and whose heights are equal to the function values of the right corner of the rectangles
Find the area $A$A under $f\left(x\right)$f(x) by partitioning $\left[0,6\right]$[0,6] into six sub-intervals of equal length and using rectangles for each sub-interval whose heights are equal to the function values of the left side of the sub-intervals.
Find the area $A$A under $f\left(x\right)$f(x) by partitioning $\left[0,6\right]$[0,6] into six sub-intervals of equal length and using rectangles for each sub-interval whose heights are equal to the function values of the right side of the sub-intervals.
What is the actual area $A$A?
Question 2
The interval $\left[0,8\right]$[0,8] is partitioned into four sub-intervals $\left[0,2\right]$[0,2], $\left[2,4\right]$[2,4], $\left[4,6\right]$[4,6], and $\left[6,8\right]$[6,8].
| $x$x | $0$0 | $2$2 | $4$4 | $6$6 | $8$8 |
|---|---|---|---|---|---|
| $y$y | $11$11 | $5$5 | $10$10 | $5$5 | $5$5 |
Approximate the area $A$A using rectangles for each sub-interval whose heights are equal to the function values of the left side of the sub-intervals.
Now approximate the area $A$A using rectangles for each sub-interval whose heights are equal to the function values of the right side of the sub-intervals.
Question 3
Approximate $\int_0^88xdx$∫808xdx by using four rectangles of equal width whose heights are the values of the function at the midpoint of each rectangle.