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16 
Evaluating the integrating constant C
Antidifferentiating then substituting in a known 'x' and 'y' value to determine the value of 'C' Finding the original equation when given a derivative and point on the original function. 
Ex16.03 NuLake IES p104 to 105 

17 
Harder finding C
Practice finding 'C' from different situations  eg. given a gradient and a point 
NuLake IES p106 to 107 
The derivative of a constant is 0  eg. 

So the antiderivative of 0 is any constant 'c'  or 

Mouse over box 

Which is the antiderivative?  
To find 'c' use a coordinate eg.(1,8) or value f(1)=8 

Steps: 1) Antidifferentiate 

2) Find 'c'  
3) Solve for 'c'  
4) Function 
Find the equation for the graph of f(x) given that f'(x) = 6x^{2} + 8 and the graph of f(x) passes through the point (1,8) 

Steps: 
f'(x) = 6x^{2} + 8 
1) Anti differentiate to find function f(x) 
f(x) = 2x^{3} + 8x + C 
2) Substitute in x = 1 & y = 8 from the coordinate and rearrange to find 'C' 
f(x) = 2x^{3} + 8x + C 
3) Write out the equation for f(x) 
f(x) = 2x^{3} + 8x + C 