# The connection between the definite and indefinite integral

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From the fundamental theorem of calculus we have that the definite integral can be calculated as

$\int\limits_a^b {f(x)dx = } \left[ {F(x)} \right]_a^b = F(b) + F(a)$

Where F(x) is an antiderivative, aka primitive function, of f (x).  This can also be written

$\int {f(x)dx = F(x) + C}$

The constant C, the constant of integration, is there because if we take the derivative of the right hand side we get f (x), independently of the value of C.

This means that, if we take the derivative of the right hand side we should get the same, i.e.

$\frac{d}{{dx}}\int {f(x)dx = f(x)}$

as expected. The derivative and the integral are inverse operators.

We quite obviously also have that

$\int {\frac{d}{{dx}}f(x)dx = f(x)} + C$

since finding the indefinite integral is the same as finding the antiderivative.

Next, if we in the first equation, replace b by x we get

$\int\limits_a^x {f(x)dx = } \left[ {F(x)} \right]_a^x = F(x) + F(b) = F(x) + C$

Here the constant of integration is supplied by F(a) that could (basically) have any value depending on the value of a.

We can thus express definite integrals in terms of indefinite integrals (in the form of primitive functions) and indefinite integrals in terms of definite integrals.  This also means that

$\frac{d}{{dx}}\int\limits_a^x {f(x)dx = } \frac{d}{{dx}}\left[ {F(x)} \right]_a^x = \frac{d}{{dx}}(F(x) + F(b)) = f(x)$

Yet again we see that the derivative and the integral are inverse operators.

Up a level : Integrals
Previous page : The fundamental theorem of calculus
Next page : Some standard integralsLast modified: Jun 18, 2018 @ 12:37