## Theorem 7.1.18: Fundamental Theorem of Calculus

Suppose

*f*is a bounded, integrable function defined on the closed, bounded interval*[a, b]*, define a new function:ThenF(x) = f(t) dt

*F*is continuous in*[a, b]*. Moreover, if*f*is also continuous, then*F*is differentiable in*(a, b)*andF'(x) = f(x)for allxin(a, b).

**Note:** In many calculus texts this theorem is called the **Second**
fundamental theorem of calculus. Those books also define a
First Fundamental Theorem of Calculus.

### Proof:

The first assumption is simple to prove: Take*x*and

*c*inside

*[a, b]*. Since

*f*is bounded we know that

*| f(x) | < M*for some number

*M*. By the properties of the Riemann integral we know:

assuming without loss of generality that

*c > x*. But then we can take the limit as

*x*approaches

*c*to see that

which implies continuity of| F(x) - F(c) | = 0

*F*at

*x = c*.

Now we want to prove that
*F(x)* is differentiable with *F' = f* if
*f* is continuous.

Pick *x* inside the interval *(a, b)* and choose a
number *h* so small that *x+h* is also in
*(a, b)*. We compute the difference quotient:

and define

Clearly we have thatm = inf{ f(t), t [x, x+h] }

M = sup{ f(t), t [x, x+h] }

*m f(t) M*so that by the properties of the Riemann integral we have that:

But since

*f*is continuous at

*x*we know that

*m = M = f(x)*as

*h*goes to zero. Therefore:

which proves the second assertion.