Calculating the area under a curve has many important applications across a wide range of subjects. As well as finding the area of shapes with curved boundaries, the area under a rate of change graph will give us the net change. For example, the area under a graph displaying speed (rate of change of distance with respect to time) would give us the total distance travelled. This concept is applicable across diverse fields such as physics, economics and biology, leading integral calculus to be a fundamental branch of mathematics.
Let's first look at calculating the area under a curve using geometric reasoning, approximating areas under curves, and the notation for such areas before we look at finding areas using calculus in the next lesson.
Areas under curves using geometric reasoning
If the area formed under a curve is a common geometric shape, such as the area formed under a straight line or semi-circle, we can use known formulas to find the exact area.
Worked examples
Example 1
Find the area between the function $f\left(x\right)=2x$f(x)=2x and the $x$x axis between $x=1$x=1 and $x=4$x=4.
The image below shows the graph $y=f\left(x\right)$y=f(x) together with the area required:
Think: Recall that the area formula for a trapezium is $A=\frac{1}{2}\left(a+b\right)h$A=12(a+b)h, where $a$a and $b$b are the lengths of the parallel sides and $h$h is the perpendicular height. Here the parallel sides are of length $2$2 and $8$8, and the perpendicular height is the length of the shape on the $x$x-axis, which is $3$3.
Do:
| Area | $=$= | $\frac{1}{2}\left(a+b\right)h$12(a+b)h |
| $=$= | $\frac{1}{2}\left(2+8\right)3$12(2+8)3 | |
| $=$= | $15$15 units2 |
Example 2
Find the area between the function $f\left(x\right)=\sqrt{4-x^2}$f(x)=√4−x2 and the $x$x-axis between $x=0$x=0 and $x=2$x=2.
The image below shows the graph $y=f\left(x\right)$y=f(x) together with the area required:
Think: Recall that the formula for the area of a circle is $A=\pi r^2$A=πr2; here we have one quarter of a circle of radius $2$2.
Do:
| Area | $=$= | $\frac{1}{4}\times\pi r^2$14×πr2 |
| $=$= | $\frac{1}{4}\times\pi\left(2\right)^2$14×π(2)2 | |
| $=$= | $\pi$π units2 |
Practice questions
Question 1
Consider the function drawn below:
Calculate geometrically, the area bounded by the curve and the $x$x-axis over $0\le x\le4$0≤x≤4.
Question 2
The function $f\left(x\right)$f(x) is defined as:
 |
$2x$2x | if $0\le x\le3$0≤x≤3 | ||
| $f\left(x\right)$f(x) | $=$= | $6$6 | if $3 |
|
| $18-2x$18−2x | if $6\le x\le9$6≤x≤9 |
Graph $f\left(x\right)$f(x) on the axis below.
Loading Graph...Hence, calculate geometrically the area bounded by the curve and the $x$x-axis.
Approximating areas under curves
If the shape formed by the curve does not resemble a common geometric shape we can use one of several methods to estimate the area. A common approach is to slice the region into rectangles and find the sum of the areas of the rectangles. Let's explore this idea.
Exploration
Consider the function $f\left(x\right)=x^2+1$f(x)=x2+1 and the area bounded by this curve and the $x$x-axis between $x=1$x=1 and $x=5$x=5.
The image below shows the graph $y=f\left(x\right)$y=f(x) together with the area of interest:
To find an approximation for this area we could divide the area into $4$4 rectangles of equal width. These rectangles could be created such that their height reached the graph at the left endpoint of each interval, the right endpoint of each interval or the mid-point of each interval. Illustrated in the graphs below.
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| Left endpoint approximation | Right endpoint approximation | Midpoint approximation |
Let's look at calculating the estimate in each case:
Case 1. Left endpoint approximation
The width of each rectangle is $1$1 and the height is the function evaluated at the left endpoint of each interval.
| Area | $\approx$≈ | $1\times f(1)+1\times f(2)+1\times f(3)+1\times f(4)$1×f(1)+1×f(2)+1×f(3)+1×f(4) |
| $=$= | $2+5+10+17$2+5+10+17 | |
| $=$= | $34$34 units2 |
We have an approximation for the area of $34$34 units2, and we can see from the graph that in this case this would be an underestimation of the actual area.
Case 2. right endpoint approximation
The width of each rectangle is $1$1 and the height is the function evaluated at the right endpoint of each interval.
| Area | $\approx$≈ | $1\times f(2)+1\times f(3)+1\times f(4)+1\times f(5)$1×f(2)+1×f(3)+1×f(4)+1×f(5) |
| $=$= | $5+10+17+26$5+10+17+26 | |
| $=$= | $58$58 units2 |
We have an approximation for the area of $58$58 units2, and we can see from the graph that in this case this would be an overestimation of the actual area.
Case 3. mid-point approximation
The width of each rectangle is $1$1 and the height is the function evaluated at the mid-point of each interval.
| Area | $\approx$≈ | $1\times f(1.5)+1\times f(2.5)+1\times f(3.5)+1\times f(4.5)$1×f(1.5)+1×f(2.5)+1×f(3.5)+1×f(4.5) |
| $=$= | $3.25+7.25+13.25+21.25$3.25+7.25+13.25+21.25 | |
| $=$= | $45$45 units2 |
We have an approximation for the area of $45$45 units2. It is not obvious from the graph in this case whether this would be an under or overestimation of the actual area.
The actual area of the given region is $45\frac{1}{3}$4513 units2, so in fact the mid-point approximation was quite close.
To obtain more accurate estimations we can use more rectangles with a reduced width to estimate the area. In the diagram below we can see that as the number of rectangles increase the amount of area underestimated is decreasing and the approximation approaches the actual area.
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 |
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| $4$4 rectangles | $8$8 rectangles | $16$16 rectangles |
Let's calculate the improved estimate for $8$8 rectangles using a left endpoint approximation:
The width of each rectangle is $0.5$0.5 and the height is the function evaluated at the left endpoint of each interval.
| Area | $\approx$≈ | $0.5\times f(1)+0.5\times f(1.5)+0.5\times f(2)+0.5\times f(2.5)+0.5\times f(3)+0.5\times f(3.5)+0.5\times f(4)+0.5\times f(4.5)$0.5×f(1)+0.5×f(1.5)+0.5×f(2)+0.5×f(2.5)+0.5×f(3)+0.5×f(3.5)+0.5×f(4)+0.5×f(4.5) |
| $=$= | $0.5\left(f(1)+f(1.5)+f(2)+f(2.5)+f(3)+f(3.5)+f(4)+f(4.5)\right)$0.5(f(1)+f(1.5)+f(2)+f(2.5)+f(3)+f(3.5)+f(4)+f(4.5)) | |
| $=$= | $0.5\left(2+3.25+5+7.25+10+13.25+17+21.25\right)$0.5(2+3.25+5+7.25+10+13.25+17+21.25) | |
| $=$= | $39.5$39.5 units2 |
We have an approximation for the area of $39.5$39.5 units2, we can see from the graph that in this case this would be an underestimation of the actual area but a closer approximation than the $34$34 units2 we calculated with only $4$4 rectangles.
Using the left endpoint approximation over an interval where a function is increasing will provide an underestimate of the area and it will provide an overestimate over an interval where a function is decreasing. This in illustrated below:
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| Left endpoint approximation where the function is increasing on the interval | Left endpoint approximation where the function is decreasing on the interval |
The opposite is true for the right endpoint approximation - it will provide an overestimate where the function is increasing on the interval and an underestimate where the function is decreasing on the interval.
These methods are not restricted to evaluating the area under a graph over an interval where the function is strictly increasing or decreasing. In our next lesson we will see how to calculate the area algebraically using integration, however, not all functions can be integrated and approximation methods such as those above are still required in such cases.
Formulas for these approximations can be useful when using many rectangles to subdivide an area. We can then evaluate these approximations using the calculator and observe the limit as the number of rectangles, $n$n, increases. The exact area can be found as the limit as $n\rightarrow\infty$n→∞.
Given a continuous function on the interval $\left[a,b\right]$[a,b], where $f\left(x\right)\ge0$f(x)≥0 for all $x$x in the interval. The area under the graph of $y=f\left(x\right)$y=f(x) from $x=a$x=a to $x=b$x=b can be approximated by subdividing the interval into $n$n rectangles of width $w=\frac{b-a}{n}$w=b−an and using one of the following methods:
Left endpoint approximation
| $A_L$AL | $=$= | $wf(a)+wf(a+w)+wf(a+2w)+wf(a+3w)\dots+wf(a+(n-1)w)$wf(a)+wf(a+w)+wf(a+2w)+wf(a+3w)…+wf(a+(n−1)w) |
| $=$= | $\sum_{k=1}^nwf(a+(k-1)w)$n∑k=1wf(a+(k−1)w) |
Right endpoint approximation
| $A_R$AR | $=$= | $wf(a+w)+wf(a+2w)+wf(a+3w)+wf(a+4w)\dots+wf(a+nw)$wf(a+w)+wf(a+2w)+wf(a+3w)+wf(a+4w)…+wf(a+nw) |
| $=$= | $\sum_{k=1}^nwf(a+kw)$n∑k=1wf(a+kw) |
Midpoint approximation
| $A_M$AM | $=$= | $wf(a+\frac{1}{2}w)+wf(a+\frac{3}{2}w)+wf(a+\frac{5}{2}w)\dots+wf(a+(n-\frac{1}{2})w)$wf(a+12w)+wf(a+32w)+wf(a+52w)…+wf(a+(n−12)w) |
| $=$= | $\sum_{k=1}^nwf(a+(k-\frac{1}{2})w)$n∑k=1wf(a+(k−12)w) |
Practice questions
Question 3
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 4
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 5
Consider the function $f\left(x\right)=10e^x$f(x)=10ex.
Which approximation method will give an underestimate of the true area on any given interval?
The right endpoint approximation because the function is always increasing.
AThe left endpoint approximation because the function is always increasing.
BThe right endpoint approximation because the function in always decreasing.
CThe left endpoint approximation because the function in always decreasing.
D
Notation
Given a continuous function on the interval $\left[a,b\right]$[a,b], where $f\left(x\right)\ge0$f(x)≥0 for all $x$x in the interval. The value of the area under the graph of $y=f\left(x\right)$y=f(x) from $x=a$x=a to $x=b$x=b is called the definite integral and is written:
That is, for a function such as $f\left(x\right)$f(x) shown in the diagram above, where $f\left(x\right)\ge0$f(x)≥0 on the interval $a\le x\le b$a≤x≤b, we have $\int_a^bf(x)dx=A$∫baf(x)dx=A.
We can see the notation includes the notation for anti-differentiation. In our next lesson, we will look at how to evaluate areas using integration and what the definite integral means for functions that lie all or partially below the $x$x-axis.
Practice questions
Question 6
Find the exact value of $\int_0^{12}f\left(x\right)dx$∫120f(x)dx geometrically, where $y=f\left(x\right)$y=f(x) is graphed below.
Question 7
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.