12.06 Definite integrals and the area under a curve

In this lesson we will show how integration can be used to calculate the exact area between the horizontal axis, a curve and two bounding values.

The integral and the area

Let's explore the concept of using infinitely many rectangles to determine the area under a curve. If we increase the number of rectangles under a curve the accuracy of the area will also increase. The following applet demonstrates this. You may notice that approximating the area by using rectangles from above and below become closer as we increase the number of rectangles.

Created with Geogebra

 

Let's consider the more generalised case as shown below. Consider some function $y=f\left(x\right)$y=f(x), and let's approximate the area of the function using $n$n rectangles, estimated from above, with all the same width. Then we can label the endpoints of the $i$ith rectangle as $x=x_{i-1}$x=xi−1​ and $x=x_i$x=xi​, where $a=x_0$a=x0​, $b=x_n$b=xn​, and $x_0x0​<x1​<x2​<…<xn​ describe the endpoints of all $n$n rectangles.

The width of one rectangle is $\Delta x$Δx, and the height of the rectangle is the function evaluated at the left-most $x$x-value of the rectangle, i.e. $f\left(x_{i-1}\right)$f(xi−1​) for the $i$ith rectangle. Hence the area of one rectangle is given by $f\left(x_{i-1}\right)\Delta x$f(xi−1​)Δx.

So for $n$n rectangles, and summing together between $a$a and $b$b we get the following approximation for the area under the function:

$f\left(x_0\right)\Delta x+f\left(x_1\right)\Delta x+f\left(x_2\right)\Delta x+\ldots+f\left(x_n\right)\Delta x$f(x0​)Δx+f(x1​)Δx+f(x2​)Δx+…+f(xn​)Δx

We can use sigma notation, i.e. $\sum_{i=1}^n$n∑i=1 to indicate that we're adding over all the rectangles. This gives us:

$\sum_{i=1}^nf\left(x_{i-1}\right)\Delta x$n∑i=1f(xi−1​)Δx

Taking the limit of this expression as $\Delta x\rightarrow0$Δx→0 and the number of rectangles, $n$n, increases, we get:

Total area under function

$=$=

$\lim_{\Delta x\rightarrow0}\ \sum_{i=1}^nf\left(x_{i-1}\right)\Delta x$limΔx→0 n∑i=1f(xi−1​)Δx

 

The process of integration represents finding the limit of the sum of the areas under a curve as$\Delta x\rightarrow0$Δx→0 between $a$a and $b$b. Including the boundary values $a$a and $b$b as a subscript and superscript respectively, we can replace the limit and sum notation with $\int_a^b$∫ba​. $\Delta x$Δx is infinitesimally small, so we use the notation $dx$dx. The area under a curve between $a$a and $b$b is given by:

 

This is called a definite integral and can be stated in words as the integral between $a$a and $b$b of $f(x)$f(x) with respect to $x$x. For simplicity, we will focus on functions that must be continuous between $a$a and $b$b and at this stage we will only consider areas entirely above the $x$x-axis. In the following sections, we will investigate how definite integrals can be calculated.

Geometric solutions to definite integrals

Integrals in the form $\int_a^b\ f(x)\ dx$∫ba​ f(x) dx can be solved using geometrical methods if the area forms a known geometrical shape.

Worked examples

Example 1

Solve $\int_1^3\ x\ dx$∫31​ x dx using geometric methods.

Think: This integral means find the area between the function $f(x)=x$f(x)=x and the $x$x axis between $x=1$x=1 and $x=3$x=3.

The image below shows this area:

The shaded area forms a trapezium with $h=2$h=2 and lengths of the parallel sides equivalent to the function values at $x=1$x=1 and $x=3$x=3 which are $1$1 and $3$3 respectively.

Do: Using the formula for the area of a trapezium:

Area	$=$=	$\frac{h}{2}(a+b)$h2​(a+b)
Area	$=$=	$\frac{2}{2}(1+3)$22​(1+3)
Area	$=$=	$4$4 units2

Therefore:

$\int_1^3\ x\ dx$∫31​ x dx

$=$=

$4$4

Example 2

Solve $\int_0^2\ \sqrt{4-x^2}\ dx$∫20​ √4−x2 dx using geometric methods.

Think: This integral means find the area between the function $f(x)=\sqrt{4-x^2}$f(x)=√4−x2 and the $x$x axis between $x=0$x=0 and $x=2$x=2.

The image below shows this area:

The function forms a semicircle with radius $2$2. The shaded area covers a quarter of the area of the circle.

Do: Using the formula for the area of a circle:

Area	$=$=	$\frac{1}{4}\pi r^2$14​πr2
Area	$=$=	$\frac{1}{4}\pi2^2$14​π22
Area	$=$=	$\pi$π units2

Therefore:

$\int_0^2\ \sqrt{4-x^2}\ dx$∫20​ √4−x2 dx

$=$=

$\pi$π

 

Note that with the methods currently available to us, we are able to solve this integral. It is good practice to consider all solution strategies available to us when solving area problems like this.

Practice questions

Question 1

Question 2

The fundamental theorem of calculus

Of course, not all areas under curves form known geometrical shapes. The process of integration provides a method to calculate the exact area between most functions and an axis. The fundamental theorem of calculus links differentiation and integration and has two parts.

The first part states that integration is the opposite of differentiation and hence implies the existence of the anti-derivative for continuous functions. The second part of the theorem links the algebraic process of anti-differentiation to the geometrical interpretation of integration. It states that an integral for a given interval can be computed by evaluating the difference in the value of the primitive function at the endpoints of the interval.

Let's look at how this can be used to calculate some areas of simple functions.

Consider the area formed by the $x$x-axis, the function $f(x)=x$f(x)=x between $x=1$x=1 and $x=3$x=3. This area is represented by the shaded region in the image below.

We solved this in example 1 using geometrical methods and obtained an area of $4$4. Let's now calculate the same area using the fundamental theorem of calculus.

In this case, our interval ranges between $x=1$x=1 and $x=3$x=3. Hence, $a=1$a=1and $b=3$b=3. Therefore the integral can be set up as follows:

Area	$=$=	$\int_a^b\ f(x)\ dx$∫ba​ f(x) dx
Area	$=$=	$\int_1^3\ x\ dx$∫31​ x dx

 

The next step is to calculate the primitive function $F(x)$F(x). We need to introduce some new notation at this stage. This involves enclosing the primitive function in square brackets and placing the values of $a$a and $b$b as a subscript and superscript respectively to the of the right square bracket.

$\int_1^3\ x\ dx$∫31​ x dx

$=$=

$\left[\frac{x^2}{2}\right]_1^3$[x22​]31​

 

The square bracket is then resolved by substituting the boundaries values and subtracting the function value at the lower boundary from the function value at the upper boundary. Note that a constant $C$C is not included in the primitive function as it will cancel out when we subtract the function values from each other.

$\int_1^3\ x\ dx$∫31​ x dx	$=$=	$\left[\frac{x^2}{2}\right]_1^3$[x22​]31​
	$=$=	$\left[\frac{3^2}{2}-\frac{1^2}{2}\right]$[322​−122​]
	$=$=	$\left[4.5-0.5\right]$[4.5−0.5]
	$=$=	$4$4 units2

 

Hence, we correctly calculated the same area as the geometrical method. While this may seem initially more complicated, it is very useful when geometrical methods can't be used.

Worked example

Example 4

Find the area enclosed by the function $f(x)=x^2+2$f(x)=x2+2 and the $x$x-axis between $x=2$x=2 and $x=3$x=3.

Think: As this is a non-linear function we will need to use integration to find the area.

Do: The image below shows the area required.

All function values are positive. therefore we can set up our integral to find the area as follows:

$Area$Area

$=$=

$\int_2^3\ x^2+2\ dx$∫32​ x2+2 dx

 

Finding the primitive function and solving:

Area	$=$=	$\left[\frac{x^3}{3}+2x\right]_2^3$[x33​+2x]32​
	$=$=	$\left[(\frac{3^3}{3}+2\times3)-(\frac{2^3}{3}+2\times2)\right]$[(333​+2×3)−(233​+2×2)] (substituting $x=2$x=2 and $x=3$x=3)
	$=$=	$\left[15-\frac{20}{3}\right]$[15−203​]
	$=$=	$\frac{25}{3}$253​ units2

Practice questions

Question 3

Question 4

Proof of part $1$1 of the fundamental theorem of calculus

The first part of the fundamental theorem of calculus says that the rate of change of the area function is the same as the function being integrated. That is:

$\frac{d}{dx}\int_a^x\ f(t)\ dt=f(x)$ddx​∫xa​ f(t) dt=f(x)

Let's look at a geometric proof of this.

Consider the area under the curve $y=f(x)$y=f(x) defined as a function $A(x)$A(x), bounded by the horizontal axis, starting at $a$a and ending at $x$x. This is represented by the blue shaded area above. That is $A(x)=\int_a^xf\left(t\right)dt$A(x)=∫xa​f(t)dt.

Now, let's consider adding an additional area, with width $h$h shown above in red. Using our area function we can define the total area (blue $+$+ red) as:

$A(x+h)$A(x+h)

The red portion can be calculated using:

$A_{red}=A(x+h)-A(x)$Ared​=A(x+h)−A(x)

As $h$h is very small, we can also approximate this area by treating it as a rectangle, hence:

$A_{red}\approx hf(x)\approx A(x+h)-A(x)$Ared​≈hf(x)≈A(x+h)−A(x)

Dividing throughout by $h$h we get:

$f(x)\approx\frac{A(x+h)-A(x)}{h}$f(x)≈A(x+h)−A(x)h​

Taking the limit of this expression as $h\rightarrow0$h→0 we get:

$f(x)=\lim_{h\rightarrow0}\frac{A(x+h)-A(x)}{h}$f(x)=limh→0A(x+h)−A(x)h​

This limit expression represents $A'(x)$A′(x) and therefore, we can write:

$f(x)=A'(x)$f(x)=A′(x)

Which is what we wanted to prove.

Practice questions

Question 5

Signed areas and integration

Up to this point, we have only considered the calculation of definite integrals in regions where all function values are positive. Let's now look at functions that have all negative values in the region of integration.

Consider the integral of $f(x)=x^2+2x-3$f(x)=x2+2x−3 between $x=-3$x=−3and $x=1$x=1:

We can see this area is completely below the horizontal axis. Setting up the integral we get:

Area

$=$=

$\int_{-3}^1\ x^2+2x-3\ dx$∫1−3​ x2+2x−3 dx

 

Finding the primitive function and solving:

Area	$=$=	$\left[\frac{x^3}{3}+x^2-3x\right]_{-3}^1$[x33​+x2−3x]1−3​
	$=$=	$\left[(\frac{1^3}{3}+1^2-3(1))-(\frac{(-3)^3}{3}+(-3)^2-3(-3))\right]$[(133​+12−3(1))−((−3)33​+(−3)2−3(−3))]
	$=$=	$\left[(\frac{1}{3}+1-3)-(-9+9+9)\right]$[(13​+1−3)−(−9+9+9)]
	$=$=	$-\frac{5}{3}-9$−53​−9
	$=$=	$-\frac{32}{3}$−323​

 

We can see that the value of the definite integral is negative, this is called the signed area. Area, however, can't be negative and is equivalent to the absolute value of the integral in this case. That is, area is $\frac{32}{3}$323​ units².

Practice questions

Question 6

Question 7