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Maths

Integration

Indefinite integration

Integration is the inverse process to differentiation. So instead of multiplying by the index and reducing it by one, we increase the index by one and divide by the new index. For example, x^2 becomes {{x^3 } \over 3} + c.

The  + \;c appears because the derivative of any constant term is zero.

c is called the (arbitrary) constant of integration. Its value can be found when appropriate additional information is provided, and this gives a particular integral.

The rule for integration is

f'(x) = ax^n  \Rightarrow f(x) = {{ax^{n + 1} } \over {n + 1}} + c

provided n \ne -1.

This can also be written in the form \int {ax^n dx = {{ax^{n + 1} } \over {n + 1}} + c} provided n \ne -1.

In general \int {f'(x)dx = f(x) + c} or \int {({{dy} \over {dx}})dx = y + c}

Here's an example.

Find the equation of the curve for which {{dy} \over {dx}} = 4x^3  + 6x^2 and which passes through the point (1, 3).

integrating gives

y = \int ( 4x^3  + 6x^2 )dx = x^4  + 2x^3  + c

substituting x = 1 and y = 3 gives 3 = 1 + 1 + c therefore c = 1

y = x^4  + 2x^3  + 1

Question

Find the equation of the curve for which {{dy} \over {dx}} = 2x + 1 and which passes through the point (2, 9).

toggle answer

Answer

integrating

y = \int {(2x + 1)dx = x^2  + x + c} substituting x = 2 and y = 9 \eqalign{
    \Rightarrow 9 & = & 4 + 2 + c  \cr 
    \Rightarrow c & = & 3  \cr 
    \Rightarrow y & = & x^2  + x + 3 \cr}

Back to Calculus index

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