It is possible to determine a function ƒ (x) from its derivative ƒ (x) by calculating the antiderivative or integral of ƒ (x), i.e.:
if dFdx = ƒ (x), then ƒ (x) = ∫ ƒ (x) dx + C 
 where C is an integration constant (see the module on Indefinite Integration).
In this module we will see how to use integration to calculate the area under a curve.
As a revision exercise, try this quiz on indefinite integration:
Quiz 1: Select the indefinite integral of ∫ (3x^{2} − 12x) dx with respect to x:
Hint: If n ≠ −1, the integral of x^{n} is x^{n + 1}(n + 1).
Explanation: To find the indefinite integral ∫ (3x^{2} − 12x) dx we use the sum rule for integrals, rewriting it as the sum of two integrals:
∫ (3x^{2} − 12x) dx  =  ∫ 3x^{2} dx − ∫ (− 12x) dx 
=  3 ∫ x^{2} dx − 12 ∫ x dx 
Using ∫ x^{n} dx = 1n+1 x^{n+1}; n ≠ −1 with n = 2 in the first integral and n = 1 in the second gives:
3 ∫ x^{2} dx − 12 ∫ x dx  =  3 × 11 + 2 x^{1+2} − 12 × 11 + 1 x^{1+1} + C 
=  33 x^{3} − 12(1 + 1) x^{2} + C = x^{3} − 14 x^{2} + C 
It can be checked that differentiation of this result gives 3x^{2} − 12x
We define the definite integral of the function ƒ(x) with respect to x from x = a to x = b to be:
b∫a ƒ (x) dx = ƒ (x) = ƒ (b) − ƒ (a) 
 where ƒ (x) is the antiderivative of ƒ(x). We call a and b the lower and upper limits of integration, respectively. The function being integrated, ƒ(x), is called the integrand. Note the minus sign!
Note: integration constants are not written in definite integrals since they always cancel in them:
b∫a ƒ (x) dx  =  ƒ (x) 
=  (ƒ (b) + C) − (ƒ (a) + C)  
=  ƒ (b) − ƒ (a) 
Example 1: Calculate the definite integral 2∫1 x^{3} dx
Solution: From the rule ∫x^{n}dx = 1n+1 x^{n+1} we have:
2∫1  =  13 + 1 x^{3+1} 
=  14 x^{4} = 14 × 2^{4} − 14 × 1^{4}  
=  14 × 16 − 14 = 4 − 14 = 154 
Exercise 1: Calculate the following definite integrals:
Solution: To calculate 3∫0 x dx use the formula:
with n = 1. This yields:
3∫0 x dx  =  11 + 1 x^{1+1} = 12x^{2} 
=  12 × (3)^{2} − 12 × (0)^{2}  
=  12 × 9 − 0 = 92 
Solution: To calculate 2∫−1 x dx use the formula for the definite integral:
with n = 1. This yields:
2∫−1 x dx  =  11 + 1 x^{1+1} = 12 x^{2} 
=  12 × (2)^{2} − 12 × (−1)^{2}  
=  12 × 4 − 12 × (+1) = 2 − 12 = 32 
Solution: To evaluate the definite integral 2∫1 (x^{2} − x) dx we rewrite it as the sum of two integrals and use ∫x^{n}dx = 1n+1 x^{n+1}, with n = 2 in the first integral and n = 1 in the second one:
2∫1 x^{2} dx − 2∫1 x dx  =  13 x^{3} − 12 x^{2} = 13 2^{3} − 13 1^{3} − ( 12 2^{2} − 12 1^{2} ) 
=  13 × 8 − 13 × 1 − ( 12 × 4 − 12 × 1 )  
=  73 − 32 = 146 − 96 = 56 
Solution: To find the integral 2∫−1 (x^{2} − x) dx, we rewrite it as the sum of two integrals and use the result of the previous part to write is as:
2∫−1x^{2} dx − 2∫−1x dx  =  13 x^{3} − 12 x^{2} = 13 2^{3} − 13 (−1)^{3} − (12 2^{2} − 12 (−1)^{2}) 
=  13 × 8 + 13 × 1 − ( 12 × 4 − 12 × 1 )  
=  93 − 32 = 3 − 32 = 32 
Click on questions to reveal their solution
The definite integral of a function ƒ(x) which lies above the xaxis can be interpreted as the area under the curve of ƒ(x). Thus the area shaded blue in the figure on the right is given by the definite integral:
b∫a ƒ (x) dx = ƒ (x) = ƒ (b) − ƒ (a)
This is demonstrated as follows: 
Consider the area, A, under the curve y = ƒ (x)in the figure on the right. If we increase the value of x by δx, then the increase in area, δA, is approximately:
δA = y δx → δAδx = y
Here we approximate the area of the thin strip by a rectangle of width δx and height y. In the limit as the strips become thin, δx → 0, this means:
dAdx = limδx → 0 δAδx = y

The function (height of the curve) is the derivative of the area and the area below the curve is an antiderivative or integral of the function.
Note: So far we have assumed that y = ƒ (x) lies above the xaxis.
Example 2: Consider the integral 3∫0 x dx. The integrand y = x (a straight line) is shown in the figure on the right. The area A underneath the line is the blue shaded triangle. The area of any triangle is half its base times the height. For the blue shaded triangle, this is:
A = 12 × 3 × 3 = 92
As expected, the integral yields the same result:
3∫0 x dx = x^{2}2 = 3^{2}2_{ } − 0^{2}2_{ } = 92 −
0
= 92

Here is a quiz on this relation between definite integrals and the area under a curve:
Quiz 2:
Select the value of the definite integral 3∫1 2 dx, shown here:
(a)6Incorrect  please try again!
(b)2Incorrect  please try again!
(c)4Correct  well done!
(d)8Incorrect  please try again!
Hint: 2 may be written as 2x^{0}, since x^{0} = 1. 
Explanation: Using 2 = 2x^{0}, the integral is:
3∫12 dx  =  2 3∫1 x^{0} dx = 2x 
=  2 × 3 − 2 × 1  
=  6 − 2 = 4 
This can be confirmed by inspecting the figure. The area under the curve between the integration limits is the area of a square of side 2, which has area 2^{2} = 4.
Example 3: Consider the two lines: y = 3 and y = −3. Let us integrate these functions in turn from x = 0 to x = 2. For y = +3: 2∫0 (+3) dx = 3x = 3 × 2 − 3 × 0 = 6  and 6 is indeed the area of the rectangle of height 3 and length 2. However, for y = −3: 2∫0 (−3) dx = −3x = −3 × 2 − (−3 × 0) = −6 
Although both rectangles have the same area, the sign of this result is negative because the curve y = −3 lies below the xaxis. This indicates the sign convention:
If a function lies below the xaxis, its integral is negative 

If a function lies above the xaxis, its integral is positive 
Exercise 2: From the figure on the right, what can you say about the signs of the following definite integrals?
(a) B∫A y_{1} dx
Solution: The sign of the definite integral, B∫A y_{1} dx, must be negative. This is because the function y_{1}(x) is negative for all values of x between A and B. The area is all below the xaxis.
(b) D∫B y_{1} dx
Solution: The sign of the definite integral, D∫B y_{1} dx, must be positive. This is because, between the integration limits B and D, there is more area above the xaxis than below it.
(c) O∫A y_{2} dx
Solution: The sign of the definite integral, O∫A y_{2} dx, must be positive. This is because, between the integration limits A and O, there is more area above the xaxis than below it.
(d) D∫C y_{2} dx
Solution: The sign of the definite integral, D∫C y_{2} dx, must be negative. This is because, between the integration limits C and D, the integrand y_{2}(x) is always negative. Click on questions to reveal their solution 
Example 4: To calculate −2∫−6 6x^{2} dx, use ∫ax^{n}dx = an+1 x^{n+1}. Thus:
−2∫−6 6x^{2} dx  =  62 + 1 x^{2+1} 
=  63 x^{3} = 2x^{3}  
=  2 × (−2)^{3} − 2 × (−4)^{3} = −16 + 128 = 112_{ } 
Note: Even though the integration range is for negative x (from −4 to −2), the integrand, ƒ(x) = 6x^{2}, is a positive function. The definite integral of a positive function is positive. Similarly, it is negative for a negative function.
Quiz 3: Select the definite integral of y = 5x^{4} with respect to x if the lower limit of the integral is x = −2 and the upper limit is x = −1:
Explanation: The definite integral of y = 5x^{4} with respect to x if the lower limit of the integral is x = −2 and the upper limit is x = −1 can be written as:
From the basic result B ∫Aax^{n}dx = an+1 x^{n+1} we obtain:
−1∫−2 5x^{4} dx  =  55 x^{5} 
=  (−1)^{5} − (−2)^{5}  
=  −1 − (−32)  
=  −1 + 32 = 31 
Note: Since the integrand 5x^{4} is positive for all x, the negative suggested solutions could not be correct.
Exercise 3: Use the integrals listed below to calculate the following definite integrals:
ƒ (x)  x^{n} for n ≠ −1  sin (ax)  cos (ax)  e^{ax}  1x 

∫ ƒ (x) dx  1n + 1 x^{n+1}  − 1a cos (ax)  1a sin (ax)  1a e^{ax}  ln (x) 
Solution: To calculate the definite integral 9 ∫4 3 √t dt we rewrite it as:
and use ∫x^{n}dx = 1n+1 x^{n+1}, with n = ½:
3 × 9 ∫4 t^{½} dt  =  1½ + 1 t^{½+1} = 3 × 1 3 ⁄ 2 t^{3 ⁄ 2} = 3 × 2 3 t^{3 ⁄ 2} 
=  2t^{3 ⁄ 2}  
=  2 × (9)^{3 ⁄ 2} − 2 × (4)^{3 ⁄ 2} = 2 × (9^{½})^{3} − (4^{½})^{3}  
=  2 × (9)^{3 ⁄ 2} − 2 × (4)^{3 ⁄ 2} = 2 × 3^{3} − 2 × (2)^{3} = 2 × 27 − 2 × 8 = 54 − 16 = 38_{ } 
Note: Dividing by a fraction is equivalent to multiplying by its inverse (see the module on Fractions).
Solution: To calculate the definite integral 1 ∫−1 (x^{2} − 2x + 4) dx we rewrite it as a sum of integrals:
1 ∫−1 x^{2} dx − 2 × 1 ∫−1 x dx + 4 × 1 ∫−1 dx, and use ∫x^{n}dx = 1n+1 x^{n+1}, with n = 2 in the first integral:
with n = 1 in the second integral:
and with n = 0 in the second integral:
Summing up these numbers we obtain 2 ⁄ 3 + 0 + 8 = 26 ⁄ 3.
Solution: To calculate the definite integral π∫0 sin (x) dx we note from the table that:
This yields (with a = 1):
π∫0 sin (x) dx  =  − cos (x) 
=  − (cos (π) − cos (0)) = − ((−1) − 1) = 2 
Note: It is worth emphasizing that the angles in calculus formulae for trigonometric functions are measured in radians.
Solution: To calculate the definite integral 3∫0 4 e^{2x} dx rewrite it as:
and use from the table:
This gives, for a = 2:
4 × 3∫0 e^{2x} dx  =  4 × 12 e^{2x} = 2 e^{2x} 
=  2 e^{(2×3)} − 2 e^{(2×0)} = 2 e^{6} − 2 e^{0} = 2 e^{6} − 2 ×1 = 2 e^{6} − 2 
Solution: To evaluate the definite integral 2 ∫1 3t dt we write:
and use:
This yields:
3 × 2 ∫1 1t dt  =  3 × ln (t) 
=  3 × ln (2) − 3 × ln (1)  
=  3 × ln (2) − 3 × 0  
=  3 ln (2) 
Note: ln (0) = 1, since e^{0} = 1.
Solution: To find the definite integral π ⁄ 2 ∫π ⁄ 4 2 cos (4w)dw use
and
2 × π ⁄ 2 ∫π ⁄ 4 cos (4w)dw  =  2 × 14 sin (4w) = 12 sin (4w) 
=  12 sin (4 × π2) − 12 sin (4 × π4)  
=  12 sin (2π) − 12 sin (π)  
=  12 × 0 − 12 × 0 = 0 
Click on questions to reveal their solution
Quiz 4: Which is the correct result for the definite integral 2b ∫ax^{2} dx?
Explanation: To caclulate the definite integral 2b∫a x^{2} dx use the basic indefinite integral:
with n = 2. This gives:
2b∫a x^{2} dx  =  12 + 1 x^{(2+1)} = 13x^{3} 
=  13 × (2b)^{3} − 13 × 1(a)^{3}  
=  13 × (2)^{3} × b^{3} − 13 × a^{3}  
=  13 × 8 × b^{3} − 13 × a^{3}  
=  83b^{3} − 13a^{3} 
Quiz 5: Which is the correct answer for the definite integral 3∫2 1x^{2} dx?
Explanation: To evaluate the definite integral 3∫2 1x^{2} dx = 3∫2 x^{−2} dx
use B∫A x^{n}dx = 1n+1 x^{n+1} with n = −2:
3∫2 x^{−2} dx  =  1−2 + 1 x^{(−2+1)} = 1(−1)x^{−1} 
=  (−1) × 1x = − 1x = − 13 − (12)  
=  − 13 + 12 = − 26 + 36  
=  −2 + 36 =16 
Choose the solutions from the options given