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 3 cases that we will consider here, the area between:
- a graph and the x axis
- a graph and the y axis
- two graphs
On this page, you should learn about
- area of a region enclosed by a curve and the x axis
- area of a region enclosed by a curve and the y axis
- area between two graphs
Here are all the different cases that you need to know about
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:
\(\int _{ a }^{ b }{ f(x) } \ dx\)
The area between the curve x=f(y) and the y axis can be found using the formula
\(\int _{ a }^{ b }{ x } \ dy\)
or using function notation:
\(\int _{ a }^{ b }{ f(y) } \ dy\)
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.
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
Print from here
Here are quizzes that will help you practise the skills from this page
START QUIZ!
Which of the following calculates the shaded area. Select ALL correct answers
There are 2 correct ways of finding the area of this region.
- You can find the area between the curve and the y axis \(\int _{ a }^{ b }{ xdy } \)
- You can subtract the area between the curve and the x axis from the area of 4 by 2 rectangle
The two functions, f and g are
f(x) = cos²x
g(x) = cos2x
Which one of the following will give the area of the region shaded below
f(x) = cos²x is the upper curve
We need to find the area under f and then subtract the area under g
Let \(f(x)=e^{x-1}\)
The diagram below shows the graph of y = f(x) and the tangent at the point x = 1
Which of the following gives the area of the shaded region
We need to find the area under the curve and subtract the area under the tangent (the triangle).
The area under the curve = \(\int _{ 0 }^{ 1 }{ f(x)dx }\)
The height of the triangle is \(f(1)=e^0=1\)
The area of the triangle = \(\frac{1\times1}{2}=0.5\)
Hence the shaded area = \(\int _{ 0 }^{ 1 }{ f(x)dx } -0.5\)
The diagram shows the graph of y = f(x).
The area of the shaded region = 3.75.
What is the value of \(\int _{ 1 }^{ 8 }{ { f }^{ -1 }(y)dy } \)
The question is asking us to find the area between the curve and the y axis shown below
To find this area, start with the area formed by the 8 by 2 rectangle, subtract the 1 by 1 rectangle and subtract the area given in the question
Find the area of the region bounded by the graph \(y = \frac{1}{x}\), x = 1 , x = e and the x axis
Area =
The area of the region can be found by working out the following integral
\(\int _{ 1 }^{ e }{ \frac { 1 }{ x } dx={ [ln|x|] }_{ 1 }^{ e } } \)
= lne - ln1
= 1 - 0
=1
The area of the region bounded by the graph \(y = \frac{1}{2x}\), x = a, x = 1 and the x axis is equal to \(ln3\)
Work out the value of a
a =
\(\int _{ 1 }^{ a }{ \frac { 1 }{ 2x } dx } \)
\(=\frac { 1 }{ 2 } \int _{ 1 }^{ a }{ \frac { 1 }{ x } dx } \)
\(=\frac { 1 }{ 2 } { [ln|x|] }_{ 1 }^{ a }\)
\(=\frac { 1 }{ 2 } (lna-ln1)\)
\(=\frac { 1 }{ 2 } lna\)
\(=ln{ a }^{ \frac { 1 }{ 2 } }\)
\(ln{ a }^{ \frac { 1 }{ 2 } }=ln3 \\ { a }^{ \frac { 1 }{ 2 } }=3 \\a=9\)
The diagram below shows the graphs of y = x² and y = 2 - x²
The area of the shaded region = \(\frac{a}{3}\)
Find the value of a
a =
Area = \(\int _{ -1 }^{ 1 }{ (2-x²)dx } -\int _{ -1 }^{ 1 }{ x²dx } \)
We can put the two integrals together
Area = \(\int _{ -1 }^{ 1 }{ (2-x²-x²)dx } \)
\(=\int _{ -1 }^{ 1 }{ (2-2x²)dx } \)
\(={ \left[ 2x-\frac { 2{ x }^{ 3 } }{ 3 } \right] }_{ -1 }^{ 1 }\)
\(=\left[ 2-\frac { 2 }{ 3 } \right] -\left[ -2+\frac { 2 }{ 3 } \right] \)
\(=\frac{8}{3}\)
The diagram below shows the graphs of y = sinx and y = cosx.
The area of the shaded region = \(a\sqrt{2}\)
Find the value of a
a =
Area = \(\int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ sinxdx } -\int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ cosxdx } \)
We can put the two integrals together
Area = \(\int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ (sinx-cosx)dx } \)
\(={ \left[ -cosx-sinx \right] }_{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }\)
\(=\left[ -cos\frac { 5\pi }{ 4 } -sin\frac { 5\pi }{ 4 } \right] -\left[ -cos\frac { \pi }{ 4 } -sin\frac { \pi }{ 4 } \right] \)
\(=\left[ \frac { \sqrt { 2 } }{ 2 } +\frac { \sqrt { 2 } }{ 2 } \right] -\left[ -\frac { \sqrt { 2 } }{ 2 } -\frac { \sqrt { 2 } }{ 2 } \right] \)
\(=\left[ \sqrt { 2 } \right] -\left[ -\sqrt { 2 } \right] \)
\(=2\sqrt { 2 } \)
Let f(x) = sinx and g(x) = sin2x
The graphs of y = f(x) and y = g(x) meet at x = 0 and x = \(\frac{\pi}{3}\) as shown below.
Find the area of the shaded region
Area =
Area = \(\int _{ 0 }^{ \frac { \pi }{ 3 } }{ (sin2x-sinx)dx } \)
\(={ \left[ -\frac { cos2x }{ 2 } +cosx \right] }_{ 0 }^{ \frac { \pi }{ 3 } }\)
\(=\left[ -\frac { cos\frac { 2\pi }{ 3 } }{ 2 } +cos\frac { \pi }{ 3 } \right] -\left[ -\frac { cos0 }{ 2 } +cos0 \right] \)
\(=\left[ \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \right] -\left[ -\frac { 1 }{ 2 } +1 \right] \)
\(=\frac { 1 }{ 4 } \)
The diagram below shows the graph of \(y = \frac{2x}{x^2-1}\)
The shaded region has an area \(lna\)
Find the value of a
a =
Area = \(\int _{ 2 }^{ 4 }{ \frac { 2x }{ x²-1 } dx } \)
We might recognise that this integral is in the form \(=\int { \frac { f'(x) }{ f(x) } dx=ln|f(x)|+C } \)
Or, we could use Integration by substitution with the substitution u = x² - 1
\(\int _{ 2 }^{ 4 }{ \frac { 2x }{ x²-1 } dx } ={ \left[ ln|x²-1| \right] }_{ 2 }^{ 4 }\)
= ln 15 - ln 3
\(=ln\frac { 15 }{ 3 } \)
= ln3
START QUIZ!
The function y = f(x) is plotted below
The area bounded by the function y = f(x), the x axis, x = -1 and x = 2 is given by
\(\int _{ b }^{ a }{ f(x)dx } \quad +\quad \left| \int _{ d }^{ c }{ f(x)dx } \right| \)
Find a , b , c & d
a =
b =
c =
d =
The important thing about this question is that the absolute symbol around the 2nd integral makes the result a positive value.
Answers c and d could be reversed.
The graph below shows the area bounded by y = x² and y = \(\sqrt{x}\)
Which expression represents the area?
The area between the graph y = lnx, the x axis and the lines x = \(\frac{1}{e}\)and x = e is in the form \(2-\frac{a}{e}\)
Find a
This question requires the integral of lnx. This is found using integration by parts. See 
integration of lnx in "special integrals"
Area = \(\left| \int _{ \frac { 1 }{ e } }^{ 1 }{ lnx }\ dx \right| \quad +\quad \int _{ 1 }^{ e }{ lnx } \ dx\)
= \({ \left| { [xlnx-x] }_{ \frac { 1 }{ e } }^{ 1 } \right| \quad +\quad \left[ xlnx\quad -\quad x \right] }_{ 1 }^{ e }\)
= \({ \left| { [1ln1-1]\quad -\quad [\frac { 1 }{ e } ln(\frac { 1 }{ e } )-\frac { 1 }{ e } ] } \right| \quad +\quad ([elne\quad -\quad e]-[1ln1\quad -\quad 1]) }\)
= \({ \left| { [0-1]\quad -\quad [\frac { 1 }{ e } (-1)-\frac { 1 }{ e } ] } \right| \quad +\quad ([e(1)\quad -\quad e]-[0\quad -\quad 1]) }\)
= \({ \left| { [-1] - [-\frac { 2 }{ e } ] } \right| \quad +\quad ([0]-[- 1]) }\)
= \({ \left| { -1 +\frac { 2 }{ e } } \right| \quad +\quad 1 }\)
= \({ { 1 -\frac { 2 }{ e } } \quad +\quad 1 }\)
= \({ { 2 -\frac { 2 }{ e } } }\)
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
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
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
The graph below shows the function f(x) = sinx where a < b \(\le \frac { \pi }{ 2 } \)
a) Write an expression for the area A, bounded by the curve y = f(x), the x axis, x = a and x = b
b) Hence, without using any further integration, show that the area B = \(bsinb - asina - cosb + cosa\)
c) Area B can also be found by finding the area bounded by the curve x = f(y), the y axis, y = f(a) and y = f(b). Find the area B using this method and show that your answer is the same as the one you found in part b)
Hint
Full Solution
How much of Area between Graphs HL have you understood?
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