1. Anti-differentiation
(integration)
mx241

2. Remember how to differentiate?
If f(x) = axn then f’(x) = naxn-1
The reverse process of differentiation is called anti-differentiation
Given f’(x) = 18x2, what would you do to get back to the original function f(x)?

3. So to anti-differentiate Raise the exponent of x by 1 axn → axn+1
To differentiate
First multiply the coefficient of x by the exponent
axn → naxn
Then reduce the exponent of x by 1
→ naxn-1
So to anti-differentiate
Raise the exponent of x by 1
axn → axn+1
Then divide by the new numerical exponent of x → a xn+1
n+1
MX241 pg 22

4. For f ’(x) = 18x2
f(x) = x2+1
2+1
= 18 x3 3
= 6x3
Check by differentiating
f(x) = 6x3
f ’(x) = 18x2

5. Differentiate the following
f(x) = x2
f(x) = x2 + 1
f(x) = x2 – 7
f(x) = x2 + c
What do you notice about all your answers?
Write down another function h(x) for which h ‘ (x) = 2x
Copy and complete this statement:
If f ‘ (x) = 2x then f(x) = ………….where c is a constant

6. This shows there is not just one answer to an anti-differentiation question.
The constant term in a function does not affect the gradient of the curve; it just translates the curve up or down the y-axis.

7. If f ‘(x) = axn then f(x) = a xn+1 + c n+1
If f ’(x) = 5x2, find the anti-derivative
5x2 → 5x3 …raise the power of x by 1
5x3 → 5 x3 …divide by the new power 3
5x3 → 5 x3 + c …include the constant c
f(x) = 3 x 5x3-1 …check by differentiation
= 5x2

8. Two special cases
If m is the gradient, then the equation of the straight line is f(x) = mx + c
therefore the anti-derivative of f ’(x) = m is f(x) = mx + c
If the gradient is zero, then the equation of the line is f(x) = c
therefore the anti-derivative of f ‘(x) = 0 is f(x) = c

9. Anti-differentiate the following functions:
f ’(x) = 0
h ’(x) = -7
t ’(x) = 3x
p ’(x) = k
f(x) = c
h(x) = -7x + c
t(x) = 1 ½ x2 + c
p(x) = kx + c

10. Integration Anti-differentiation is often shortened to integration
∫ f(x) dx
Means integrate the function f(x) with respect to x
Note: dx has a single symbol meaning; it does not mean d times x

11. Integration of polynomials
Find ∫ (2x + 1)(x – 3) dx
∫ (2x + 1)(x – 3) = ∫ (2x2 – 5x – 3) dx
= 2 x x2 – 3x + c

12. Integration with different variables
Find ∫ (at2 + bt + c) dt
∫ (at2 + bt + c) dt = a t3 + b t2 + ct + k
= 1 at3 + 1 bt2 + ct + k

13. Find the following integrals
∫ (3x2 + 6x – 4) dx
∫ (x2 – 3x + 2) dx
∫ (3x – 2)(x + 4) dx
∫ (t2 – 2) dt
∫ (k – pv) dv
∫ (m2x2 – 4p2) dx
= 3 x3 + 6 x2 – 4x + c
= x3 + 3x2 – 4x + c
= 1/3 x3 – 3/2 x2 + 2x + c
= ∫ 3x2 + 10x – 8) dx
= x3 + 5x2 – 8x + c
= 1/3 t3 – 2t + c
= kv – (pv)2/2 + c
= kv - (pv2)/2 + c
MX241 page 28 ex 6A
= m2/3 x3 – 4p2 x + c
= 1/3 m2 x3 – 4p2 x + c

14. If f‘(x) = , what is f(x)? f ‘ (x) = 6x-2
6x -2 → 6x -2+1 = 6x -1 …raise the power of x by 1
6x -1 → 6 x -1 …divide by the new power -1
f(x) = -6 x -1 + c …add the constant c
= c

15. Find the following integrals
∫ dx
∫ dx
= ∫ x -3 dx =
=∫ ½ x-2 dx
= x -1 + c =
= ∫ 5x -4 dx =
MX241 page 29
= ∫ ( x2 – 4x -2) dx
= x3 – =

16. Applications of anti-differentiation
Find the equation of the curve which cuts the y-axis at (0,-4) if its derived function is f ‘(x) = (x+1)2
If f ‘(x) = (x+1)2
= x2 + 2x + 1
then f(x) =
As the curve passes through the point (0,-4)
-4 = c
c = so f(x) =

17. exercises f ’(x) = 2x – 1 f(x) = x2 – x + c f ‘(x) = -6x2 + 4x + 2
g ‘(x) = (x – 3)(x + 5)
= x2 + 2x – 15
So g(x) =
Since the curve passes through (3,-5).
-5 = c
C = 22
The equation of the curve is
f ’(x) = 2x – 1
f(x) = x2 – x + c
Since the curve passes through (1,-2)
-2 = 1 – 1 + c
C = -2
The equation of the curve is f(x) = x2 – x – 2
A curve passes through the point (1,-2) is given by the function f ‘(x) = 2x – 1.
What is the equation of this curve?
2. The gradient function of a curve is g ‘(x) = (x – 3)(x + 5) and the curve passes through the point (3 , -5).
Find the equation of the curve.
3. The gradient at any point of a curve which passes through the origin is given by the function f ‘(x) = -6x2 + 4x + 2.
what is the equation of the curve?
f ‘(x) = -6x2 + 4x + 2
f(x) = -2x3 + 2x2 + 2x + c
Since the curve passes through (0,0).
0 = 0 + c
C = 0
The equation of the curve is f(x) = -2x3 + 2x2 + 2x

18. Definite Integrals
Suppose you name the anti-derivative of the function f(x) as F(x) where F’(x) = f(x)
Then you can define the definite integral as =
a is called the lower limit and b is called the upper limit of the definite integral.
This gives a numerical value that does not include the constant c.

19. Evaluate
… integrate the function (don’t include the constant) and enclose in square brackets, copying the upper and lower limits
…evaluate: (substitution of upper limit) - (substitution of lower limit).
= 16 – 1
=15

20. Evaluate (why not put in ‘c’ when solving a definite integral?)
= -3

21. Evaluate the following definite integrals
2x dx (4x + 3) dx
x3 dx (6 – x) dx
5. (4x + x2) dx (x ) dx

22. Areas under curves
The area under the curve is the area enclosed by the curve, x-axis, and the lines x=a and x=b y
y = f(x)
a b x

23. A(1) equals the area under the function
A(3) minus
A(1) equals the area under the function
The area of the function under the curve between x = 3 and x = 1 is A(3) – A(1)

24. Area by integration If the equation of the curve is y=f(x), then y
a b x

25. Find the area under the curve f(x)=x+1 between x=2 and x=5.
Here, a=2, b=5 and f(x)=x+1
Can go to theta chapter 16 ppt G, H, I and J showing the areas above, below and combined curves