REVISION NOTES
IGCSE Edexcel Further Pure Mathematics
1.9 Calculus
1.9.1 Differentiation and integration of sums of multiples of powers of x (excluding integration of 1/x),sin ax,cos ax,eax
![edexcel_igcse_further pure mathematics_topic 9_calculus_002_basic differentiation.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_002_basic-differentiation-1.png)
![edexcel_igcse_further pure mathematics_topic 9_calculus_003_basic integration.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_003_basic-integration-1.png)
1.9.2 Differentiation of a product, quotient and simple cases of a function of a function
![edexcel_igcse_further pure mathematics_topic 9_calculus_004_differentiation chain, product, quotient rule.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_004_differentiation-chain-product-quotient-rule-1.png)
1.9.3 Applications to simple linear kinematics and to determination of areas and volumes
Type 1: Area between a curve and x-axis (y > 0)
![edexcel_igcse_further pure mathematics_topic 9_calculus_009_area above x axis between curve and x axis.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_009_area-above-x-axis-between-curve-and-x-axis.png)
Type 2: Area between a curve and x-axis (y < 0)
![edexcel_igcse_further pure mathematics_topic 9_calculus_010_area below x axis between curve and x axis.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_010_area-below-x-axis-between-curve-and-x-axis-1.png)
Type 3: Area between a curve and x-axis (-∞ < y <∞)
![edexcel_igcse_further pure mathematics_topic 9_calculus_011_area above and below x axis between curve and x axis.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_011_area-above-and-below-x-axis-between-curve-and-x-axis.png)
Type 4: Area between two curves
![edexcel_igcse_further pure mathematics_topic 9_calculus_012_area between a curve and a straight line.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_012_area-between-a-curve-and-a-straight-line.png)
1.9.4 Stationary points and turning points
Coordinates of a Stationary Point
Step 1: Differentiate the equation and equate to 0 (f'(x) = 0)
Step 2: Substitute the value of x into the equation to find y
![edexcel_igcse_further pure mathematics_topic 9_calculus_007_turning point and gradient on graph.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_007_turning-point-and-gradient-on-graph-1.png)
1.9.5 Maxima and minima
![edexcel_igcse_further pure mathematics_topic 9_calculus_005_double differentiation maxima or minima.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_005_double-differentiation-maxima-or-minima-1.png)
1.9.6 The equations of tangents and normals to the curve y = f(x)
![edexcel_igcse_further pure mathematics_topic 9_calculus_001_tangent to a curve diagram.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_001_tangent-to-a-curve-diagram-1.png)
![edexcel_igcse_further pure mathematics_topic 9_calculus_006_normal to a curve.png](https://resource.studiaacademy.com/wp-content/uploads/2023/08/edexcel_igcse_further-pure-mathematics_topic-9_calculus_006_normal-to-a-curve-1.png)
1.9.7 Application of calculus to rates of change and connected rates of change
Finding rate of change of a part of a usually 3D shape (e.g. radius):
- Eg: Area of volume of cylinder info
- 50cm3/s (rate of sand poured)
- V of cone increases in a way that r of base is always 3 times the h of the cone
- Find rate of change of radius of cone, when radius is 10cm
Working:
dV/dt = 50 (given)
dr/dt = need to find
dr/dt = dV/dt x dr/dV
r = 3h (given)
h = r/3
V = 1/3πr2h
V = 1/3πr2(r/3)
V = 1/9πr3
dV/dr = 1/3πr2
dr/dt = 50 x 1/(1/3π(10)2) = 0.477 cm/s
dr/dt = 50 x 1/(1/3π(10)2) = 0.477 cm/s