Maths Learning Service: Revision
Mathematics IA
Anti-differentiation
(Integration)
Anti-differentiation
Anti-differentiation or integration is the reverse process to differentiation. For example, if
f 0 (x) = 2x, we know that this is the derivative of f (x) = x2 . Could there be any other
possible answers?
If we shift the parabola f (x) = x2 by sliding it up or down vertically, all the points on the
curve will still have the same tangent slopes, i.e. derivatives. For example:
y = x 2+ 3
y = x 2+ 1
y
y = x2 2
x
all have the same derivative function, y 0 = 2x, so a general expression for this family of curves
would be
y = x2 + c
where c is an arbitrary constant (called the integration constant).
Note: Where possible, check your answer by differentiating, remembering that the derivative
of a constant, c, is zero.
In mathematical notation, this anti-derivative is written as
Z
2x dx = x2 + c
Z
The integration symbol “ ” is an extended S for “summation”. (You will see why in Mathematics IM.)
The “dx” part indicates that the integration is with respect to x. For instance, the integral
Z
2x dt can not be found, unless x can be rewritten as some function of t.
1
Examples: (1) If y 0 = x, then y = x2 + c
2
1
(2) If y 0 = x2 , then y = x3 + c
3
(Check: y 0 =
(Check: y 0 =
1
× 2x + 0 = x
2
1
× 3x2 + 0 = x2
3
)
)
Integration
(3)
2007 Mathematics IA Revision/2
Z
x−2
x−3 dx = −
+c
2
Z
2x 2
x2
+c
+c=
x dx =
3
3/2
d
x−2
x−3
(Check:
−
+ c = −2 × −
+ 0 = x−3
dx
2
2
!
3
(4)

3

3
)
1
1
3 2x 2
d  2x 2
+ 0 = x2
+ c = ×
(Check:
3
3
2
dx
1
2
)
Notice that a pattern emerges which can be summarized in mathematical notation as
xn+1
+c
n+1
for any real number n, except −1.
Z
xn dx =
Z
x0
+ c, which is undefined. However, the
When n = −1 this formula would give x−1 dx =
Z0
d
1
integral does exist. Since
(ln x) = we can say x−1 dx = ln |x| + c.
dx
x
As a consequence of other basic rules of differentiation, we also have
Z
Z
kg(x) dx
= k
(g(x) + h(x)) dx
Z
Examples: (1)
Z
=
eax+b dx
Z
g(x) dx, where k is a constant.
g(x) dx +
Z
h(x) dx
1 ax+b
e
+ c, where a, b and c are constants.
a
=
x3 x2
+
+c
3
2
1
e3x−2 dx = e3x−2 + c
3
x2 + x dx =
Z 2x 2
8x 2
4x + 3 dx = 4 ×
+ 3x + c =
+ 3x + c
3
3
(2)
3
(3)
Z
Z 1
2
3
Exercises
1. Find the following integrals. Check each answer by differentiating.
(a)
Z
x9 dx
(b)
Z
1
x 4 dx
(c)
Z
x−5 dx
(d)
Z
3
x− 2 dx
(e)
Z
1 dx
2. Find the following integrals. Check each answer by differentiating.
Z Z Z
1
3
(a)
2x 2 + 2 + 1 dx
(b)
1 − 4x + 9x2 dx
(c) (2x + 1)2 dx
x
(d)
Z
3
dx
x2
(e)
Z
e7x dx
(f)
Z
e−x−1 dx
Integration
2007 Mathematics IA Revision/3
Definite Integration and areas under curves
The definite integral
Z
b
f (x)dx is the number F (b) − F (a), where F (x) =
Z
f (x)dx, the
a
antiderivative of f (x).
Example:
Z
2
0
x3 x2
x − x dx =
−
+c
3
2
2
"
#2
0
8 4
0 0
2
=
− +c −
− +c = .
3 2
3 2
3
Note: The constant c cancels out in definite integration.
If f (x) ≥ 0 and continuous in the interval a ≤ x ≤ b then
the curve between a and b:
Z
b
f (x)dx is the shaded area under
a
f(x)
b
a
Example:
Z
2
h
(2x + 2)dx = x2 + 2x
i2
−1
−1
x
= (4 + 4) − (1 − 2) = 9.
For this simple curve we can check the area:
y
f(x) = 22x+32
6
x
-1
2
1
×3×6=9 .
2
The area enclosed between two curves f (x) and g(x), where f (x) ≥ g(x) in the interval
a ≤ x ≤ b, is given by
The shaded area is
Z
b
a
f (x)dx −
Z
b
g(x)dx =
a
Z
b
(f (x) − g(x)) dx
a
irrespective of the position of the x−axis.
Example: Find the area enclosed between f (x) = x + 2 and g(x) = x2 + x − 2.
The curves intersect when x + 2 = x2 + x − 2 or x2 − 4 = (x − 2)(x + 2) = 0, ie. when x = ±2.
Since f (x) ≥ g(x) in the interval −2 ≤ x ≤ 2 we have
Integration
Z
2
2007 Mathematics IA Revision/4
2
x + 2 − (x + x − 2) dx
=
−2
Z
2
y
4−x
2
dx
−2
4
#2
x3
= 4x −
3
−2
8
8
− −8 +
=
8−
3
3
16
32
= 16 −
=
units2
3
3
"
2
x
-1
2
Exercises
3. Calculate
(a)
Z
5
x
e dx
(b)
Z
0
2
3x dx
(c)
−2
2
9
Z
2
√
x − x dx
(d)
Z
−1
−2
4
1
2x + 2 dx
x
3
4. Find the area between the following functions and the x−axis for the indicated interval.
√
1
(b) x − x , 4 ≤ x ≤ 9
(a) x 3 , 1 ≤ x ≤ 8
5. Find the area between (a) y = x − 2 and y = 2x − x2
(b) y = x2 and x = y 2
ANSWERS
5
x10
+c
1. (a)
10
4x 4
+c
(b)
5
(c) −
x−4
+c
4
1
(d) −2x− 2 + c
3
3
2. (a) 43 x 2 − + x + c
(b) x − 2x2 + 3x3 + c
x
3
(d) − + c
(e) 17 e7x + c
(f) −e−x−1 + c
x
3. (a) 141.024
4. (a) 11 41
5. (a) 4 21
(c) −39 23
(b) 8
(e) x + c
(c) 43 x3 + 2x2 + x + c
(d) −7
(b) 19 56
(b)
y
1
3
y
x
(2,0)
1
(1,1)
(-1,-3)
(0,0)
1
x