Area between Graphs SL

On this page, we'll look at how we can use integration to find areas. This is a really common question in examinations.  The technique is not too hard, but there are just a couple of pitfalls to avoid which will be explained. There are 2 cases that we will consider here, the area between:

  •  a graph and the x axis
  • two graphs

Key Concepts

On this page, you should learn about

  • area of a region enclosed by a curve and the x axis
  • area between two graphs

Essentials

Here are all the different cases that you need to know about

Area between Curve and x-axis

The area between the curve y=f(x) and the x axis can be found using the formula

\(\int _{ a }^{ b }{ y } \ dx\)

or using function notation:

\(\int _{ a }^{ b }{ f(x) } \ dx\)

Areas under Axes

We should always draw a sketch of the graph before we calculate the area since any part below the axis will be calculated as a negative value. Below, we consider the area between the graph y = x²-1 and the x axis. You can check these definite integrals with your GDC:

\(\int _{ 0.5}^{ 1 }{ x^2 } \ dx \approx -0.21\)

The area \(\approx 0.21\)

\(\int _{ 1}^{ 1.5 }{ x^2 } \ dx \approx 0.29\)

The area \(\approx 0.29\)

\(\int _{ 0.5}^{ 1.5}{ x^2 } \ dx \approx 0.083\)

The area is NOT equal to 0.083

Clearly, the area of the third graph is NOT 0.083 and CANNOT be found by evaluating \(\int _{ 0.5}^{ 1.5}{ x^2 } \ dx\)    

The area \(\approx 0.21+0.29\approx 0.50\)

There are different ways of finding the area is this case. The simplest is to evaluate separately the part that is above to the part that is below the axis. We then make the part below a positive value.

In this case,

area =  \(|\int _{ 0.5}^{ 1}{ x^2 } \ dx|\quad+\quad\int _{ 1}^{ 1.5}{ x^2 } \ dx\)


In General

 The area between the curve y=f(x) and the x axis can be found using the formula

\(|\int _{ a}^{ b}{ y } \ dx|\quad+\quad|\int _{ b}^{ c}{ y } \ dx|\)

We can use function notation

\(|\int _{ a}^{ b}{ f(x) } \ dx|\quad+\quad|\int _{ b}^{ c}{ f(x) } \ dx|\)

You will need to find the zeros (y intercepts) of your graph.

Area bounded by 2 Graphs

To find the area bounded by two graphs like this,  we need to subtract one area from the other:

  equals        subtract   

\(\int _{ a }^{ b }{ f(x) } \ dx\quad-\quad \int _{ a }^{ b }{ g(x) } \ dx\)

In this case, you do not have to worry about graph being above or below the x axis!


Here are video examples to help you understand the concepts from this page

Area below x-axis

In this example, we see the importance of sketching the graph to see which part of the region lies above the x axis and which lies below. Finding the definite integral between 0 and 3 would find the WRONG answer here!

Find the area bounded by the curve y= x3 – 4x2 + 3x and the x axis from x = 0 to x = 3

Area between 2 Curves

Here is a good example that combines finding the equation of a tangent , finding the area between 2 curves and thinking about areas under straight lines as triangles.

Find the area bounded by the curve y= ex , the tangent to the curve where x = 1 and the y axis.

Summary

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Test Yourself

Here are quizzes that will help you practise the skills from this page


START QUIZ!

START QUIZ!

Exam-style Questions

Question 1

Let f(x)=sinx, for \(0\le x\le 2\pi \)

The following diagram shows the graph of f

The shaded region R is enclosed by the graph of f, the line x=a , where a<\(\pi\) and the x-axis.

The area of R is \(\left( 1-\frac { \sqrt { 3 } }{ 2 } \right) \). Find the value of b.

 

Hint

 

Full Solution

 

Question 2

Show that the area bounded by the graphs of y = f(x) and y = g(x) in the interval \(0\le x\le \pi \) is given by 2 - \(\frac { \pi }{ 2} \)

f(x) = sinx

g(x) = sin²x

Hint

Full Solution

 

Question 3

Let \(f(x)=ln\left( \frac { x }{ x-1 } \right) \) for x>1

a) Find f ' (x)

b) Hence, show that the area bounded by g(x) = \(\frac { 1 }{ x(x-1) } \) , the x axis, x = 2 and x = e is given by \(ln\left( \frac { 2e-2 }{ e } \right) \)

Hint

Full Solution

MY PROGRESS

How much of Area between Graphs SL have you understood?