Integration is sometimes called antidifferentiation, as it is the opposite process of differentiation. When we find an indefinite integral, we find a function with an arbitrary constant, C. Whereas, when we find a definite integral, we find a numerical value. Definite integration has many applications. For the examination, we use it to find displacement in questions about kinematics, probabilities with continuous random variables and it is linked to work on finding areas and volumes with integration.
On this page, you should learn about
- definite integrals
- \(\int_{a}^{b} f'(x)\: dx=\: f(b)-f(a)\)
The following videos will help you understand all the concepts from this page
In the following video, we are going to look at what definite integration is (comparing to indefinite integration) and what we use it for. We'll look at some of the many applications of definite integration.
Print from here
Here is a quiz that practises the skills from this page
START QUIZ!
\(\int _{ 1 }^{ 2 }{ x^3dx=\frac { a }{ 4 } } \)
Work out a
a =
\(\int _{ 1 }^{ 2 }{ x^{ 3 }dx }\\= { \left[ \frac { { x }^{ 4 } }{ 4 } \right] }_{ 1 }^{ 2 }\\=\left[ \frac { { 2 }^{ 4 } }{ 4 } \right] -\left[ \frac { { 1 }^{ 4 } }{ 4 } \right] \\=\frac { 16 }{ 4 } -\frac { 1 }{ 4 } \\=\frac { 15 }{ 4 } \)
Evaluate \(\int _{ 0 }^{ 4 }{ \frac { 1 }{ \sqrt { 2x+1 } } dx } \)
Answer =
\(=\int _{ 0 }^{ 4 }{ { (2x+1) }^{ -0.5 }dx } \\ ={ \left[ \frac { { (2x+1) }^{ 0.5 } }{ 2\cdot 0.5 } \right] }_{ 0 }^{ 4 }\\ ={ \left[ \sqrt { 2x+1 } \right] }_{ 0 }^{ 4 }\\ =\sqrt { 9 } -\sqrt { 1 } \\ =3-1\\ =2\)
Evaluate \(\int _{ \pi }^{ 2\pi }{ \sin { \frac { x }{ 2 } dx } } \)
Answer =
\(={ \left[ -2cos\frac { x }{ 2 } \right] }_{ \pi }^{ 2\pi }\\ =\left[ -2cos\pi \right] -\left[ -2cos\frac { \pi }{ 2 } \right] \\ =2-0\)
\(\int _{ 4 }^{ 5 }{ \left( \frac { 1 }{ x+3 } -\frac { 1 }{ x-3 } \right) } dx=ln\left( \frac { a }{ b } \right) \)
Work out a and b
a =
b =
\(={ \left[ ln(x+3)-ln(x-3) \right] }_{ 4 }^{ 5 }\\ ={ \left[ ln\left( \frac { x+3 }{ x-3 } \right) \right] }_{ 4 }^{ 5 }\\ =\left[ ln\left( \frac { 8 }{ 2 } \right) -ln\left( \frac { 7 }{ 1 } \right) \right] \\ =ln4-ln7\\ =ln\frac { 4 }{ 7 } \)
The graph below shows the function f
Evaluate the following definite integrals
- \(\int _{ 1 }^{ 3 }{ f(x)dx } \) =
- \(\int _{ -1 }^{ 0 }{ f(x)dx } \) =
- \(\int _{ -1 }^{ 4 }{ f(x)dx } \) =
definite integral is negative as it is below x axis
answer = -1 + 3 + (-0.5)
The graph below shows the function f
\(\int _{ a }^{ 7 }{ f(x)dx } =0\)
Find 2 possible values of a , a1 and a2
\(a_1\) =
\(a_2\) =
You must choose a value for a that makes the definite integral from a to 7 equal 0.
Choose a value of a, so that the area of the region below the x axis must equal the area above the x axis.
Given that \(\int _{ 1 }^{ 3 }{ f(x) } dx=6\) and \(\int _{ 1 }^{ 3 }{ g(x) } dx=4\)
find the value of the following definite integrals
- \(\int _{ 1 }^{ 3 }{ (f(x) } +g(x))dx\) =
- \(\int _{ 3 }^{ 3 }{ g(x) } dx\) =
- \(\int _{ 3 }^{ 1 }{ f(x) } dx\) =
- \(\int _{ a }^{ b }{ (f(x) } +g(x))dx=\)\(\int _{ a }^{ b }{ f(x) } dx+\int _{ a }^{ b }{ g(x) } dx\)
- \(\int _{ a }^{ a}{ g(x) } dx=0\)
- \(\int _{ a }^{ b }{ f(x) } dx=-\int _{ b }^{ a }{ f(x) } dx\)
Given that \(\int _{ 2}^{ 4 }{ f(x) } dx=10\) and \(\int _{ 0}^{ 2 }{ f(x) } dx=4\)
find the value of the following definite integrals
- \(\int _{ 0}^{ 4 }{ f(x) } dx\) =
- \(\int _{ 2}^{ 0 }{ f(x) } dx\) =
- \(\int _{ 2}^{ 4 }{2 f(x) } dx\) =
- \(\int _{ a }^{ c }{ f(x) } dx=\int _{ a }^{ b }{ f(x) } dx+\int _{ b }^{ c }{ f(x) } dx\)
- \(\int _{ a }^{ b }{ f(x) } dx=-\int _{ b }^{ a }{ f(x) } dx\)
- \(\int _{ a }^{ b }{k f(x) } dx=k\int _{ a }^{ b }{ f(x) } dx\)
Given that \(\int _{ 1}^{ 3 }{ 4f(x) } dx=20\)
find the value of the following definite integrals
- \(\int _{ 3}^{ 1 }{ f(x) } dx\) =
- \(\int _{ 3}^{ 3 }{ f(x) } dx\) =
- \(\int _{ 1}^{ 3 }{ (f(x)+2x) } dx\) =
- \(\int _{ 1}^{ 3 }{ f(x) } dx\) = 5 , therefore \(\int _{ 3}^{ 1 }{ f(x) } dx\) = -5
- \(\int _{ a }^{ a}{ f(x) } dx=0\)
- \(\int _{ 1}^{ 3 }{ f(x) } dx+\int _{ 1}^{ 3 }{ 2x } dx\)
= 5 + \({ \left[ { x }^{ 2 } \right] }_{ 1 }^{ 3 }\) = 5 + [9 - 1]
= 13
Given that \(\int _{ -2 }^{ 2 }{ f(x)dx=5 } \) and \(\int _{ -2 }^{ 0 }{ f(x)dx=3 } \) work out
a) \(\int _{ 0 }^{ 2 }{ 2f(x)dx } \) =
b) \(\int _{ 2 }^{ 4 }{ f(x-2)dx } \) =
c) \(\int _{ -2 }^{ 2 }{ (f(x)+2)dx } \) =
a) Consider the same definite integral split into two parts
\(\int _{ -2 }^{ 0 }{ f(x)dx}+\int _{ 0 }^{ 2 }{ f(x)dx } =\int _{ -2 }^{ 2 }{ f(x)dx } \\3 + 2 = 5\\\int _{ 0 }^{ 2 }{ f(x)dx }=2\)
b) f(x - 2) translates the graph of f(x) 2 units to the right. This is the same as \(\int _{ 0 }^{ 2 }{ f(x)dx } \)
c) f(x) + 2 translates the graph 2 units up. It creates a region the same as \(\int _{ -2 }^{ 2 }{ f(x)dx=5 } \) , but with a rectangle below. The area of the rectangle is 8.
Given that \(\int _{ 4 }^{ 8 }{ \frac { 1 }{ 2x-4 } dx= } ln\sqrt { a } \) , find the value of a
Hint
Full Solution
Consider a function f(x) such that \(\int _{ 0 }^{ 4 }{ f(x)dx } \) = 6
Find
a) \(\int _{ 0 }^{ 4 }{3 f(x)dx } \)
b) \(\int _{ 0 }^{ 4 }{[ f(x)+3]dx } \)
c) \(\int _{ -3 }^{ 1 }{\frac{1}{3} f(x+3)dx } \)
d) \(\int _{ 0 }^{ 4 }{ [f(x)+x]dx } \)
Hint
Full Solution
How much of Definite Integration have you understood?
Twitter 
Facebook 
LinkedIn