11.02 Evaluate areas using geometry

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 $x=3$x=3.

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 $4$4.

Do:

Area	$=$=	$\frac{1}{2}\left(a+b\right)h$12​(a+b)h
	$=$=	$\frac{1}{2}\left(2+8\right)4$12​(2+8)4
	$=$=	$20$20 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

Question 2

 

 

Riemann integral approximation

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 or the right endpoint of each interval.

Left endpoint approximation	Right endpoint 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.

 

The actual area of the given region is $45\frac{1}{3}$4513​ units².

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.

$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:

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→∞.

Scroll to the bottom of this lesson for instructions on how to approximate and evaluate the areas under curves using a CAS calculator.

 

Practice questions

Question 3

Question 4

Question 5

 

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 $\int_a^bf(x)dx=A$∫ba​f(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

Question 7

 

Using a calculator to approximate and find areas under curves

See below to work through an example of using a calculator for calculations concerning areas under curves and evaluating definite integrals.

TI Nspire