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Finding the Integrating constant 'C'

Calculus HOME | Achievement Objectives | Gradient | Gradient Functions | Differentiation | Gradient at a point | Find point with gradient | Equation of Tangent | Second Derivitive | Coordinates of Max & Min | Increasing, Decreasing Functions | Applications | Kinematics with differentiation | Antidifferentiation | Finding 'C' | Kinematics with anti-differentiation | Rates of Change | Mixed Problems | Revision

16

Evaluating the integrating constant C

Anti-differentiating 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.

Class Notes | Blank Notes

Ex16.03

NuLake IES p104 to 105

 

 
17

Harder finding C

Practice finding 'C' from different situations - eg. given a gradient and a point

Class Notes | Blank Notes

NuLake IES p106 to 107

 

 

 

The derivative of a constant is 0

eg.

So the anti-derivative of 0 is any constant 'c'

or

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Which is the anti-derivative?

To find 'c' use a coordinate eg.(1,8) or value f(1)=8

Steps: 1) Anti-differentiate

            2) Find 'c'
            3) Solve for 'c'
            4) Function

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Example:
Find the equation for the graph of f(x) given that f'(x) = 6x2 + 8
and the graph of f(x) passes through the point (1,8)

Steps:

f'(x) = 6x2 + 8

1) Anti differentiate to find function f(x)

f(x) = 2x3 + 8x + C

2) Substitute in x = 1 & y = 8 from the coordinate and rearrange to find 'C'

f(x) = 2x3 + 8x + C
8 = 2x13 + 8x1 + C
8 = 2 + 8 + C
8 = 10 + C
-2 = C

3) Write out the equation for f(x)

f(x) = 2x3 + 8x + C
f(x) = 2x3 + 8x - 2

Class Notes | Blank Notes |(harder) Class Notes | Blank Notes Screen Snaps:

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