# Fundamental Theorem of Calculus

How to use the fundamental theorem of calculus to solve definite integral problems: theorem (part 1, part 2), proofs, example, and its solution.

## Definition of Definite Integral

To find the area under the graph,

first write the sum of the area *S*_{n} (Riemann sum).

Riemann integral

As *n* → ∞,*S*_{n} becomes *S* (Riemann integral).

Then you can change this Riemann integral

into a definite integral.

(A definite integral is an integral

with its limits: *a* and *b*.)

Change the limit and the sigma part into ∫.*x* goes from *a* (lower limit) to *b* (upper limit).

Change *f*(*x*_{k}) into *f*(*x*).

And change △*x* into *dx*.

So ∫_{a}^{b} *f*(*x*) *dx* means

the area under *y* = *f*(*x*)

from *x* = *a* to *x* = *b*.

This integral is read as

'the integral, from *a* to *b*, of *f*(*x*) *dx*'.

## Theorem (Part 1)

The fundamental theorem of calculus consists of two parts.

This is part 1:

If *F*(*x*) is defined as ∫_{a}^{x} *f*(*t*) *dt*,

then *F*'(*x*) = *f*(*x*).

This part shows that

the integral of *f*(*x*) is the antiderivative of *f*(*x*).

In other words, this theorem shows that

the integration and the differentiation

are the opposite operations.

This is why

when you're solving indefinite integrals,

you're finding their antiderivatives.

## Theorem (Part 2)

This is part 2:

∫_{a}^{b} *f*(*x*) *dx* = *F*(*b*) - *F*(*a*)

This part shows how to solve a definite integral.

## Proof: Part 1

By the hypothesis,*F*(*x*) is the area under the graph in [*a*, *x*]. (gray)

Then *F*(*x* + *h*) is the area under the graph in [*a*, *x* + *h*],

which is (gray) + (blue) + (red).

(blue) = *f*(*x*)⋅*h*

(red) is the upper part of the error.

It can be (-) when (blue) is taller than the graph.

Move *F*(*x*) and (blue) to the left side,

only leaving (red) on the right side.

Absolute value both sides.

See the figure above.

(purple) is the area that makes a rectangle with (red).

So |(red)| ≤ |(red) + (purple)|.

The rectangle's area is [*f*(*x* + *h*) - *f*(*x*)]⋅*h*.

So |(red) + (purple)| = | [*f*(*x* + *h*) - *f*(*x*)]⋅*h* |.

Rewrite the inequality.

|*F*(*x* + *h*) - *F*(*x*) - *f*(*x*)⋅*h*| ≤ | [*f*(*x* + *h*) - *f*(*x*)]⋅*h* |

Then divide both sides by *h*.

Solve the left absolute value sign.

Absolute value inequalities (One variable)

As *h* → 0,

the left side becomes -|*f*(*x*) - *f*(*x*)| = 0,

and the right side becomes |*f*(*x*) - *f*(*x*)| = 0.

So, by the squeeze theorem,

the middle term goes to 0.

The limit of [*F*(*x* + *h*) - *F*(*x*)]/*h* is *F*'(*x*).

Definition of a derivative function

So *F*'(*x*) - *f*(*x*) = 0.

Move *f*(*x*) to the right side.

Then *F*'(*x*) = *f*(*x*),

which is the conclusion of the theorem.

## Proof: Part 2

Start from the conclusion of part 1: *f*(*x*) = *F*'(*x*).

Integrate (or antidifferentiate) both sides.

Then, by part 1,

the left side is ∫_{a}^{x} *f*(*t*) *dt*,

and the right side is *F*(*x*) + *C*.

Antiderivative

To find *C*,

put *a* into *x*.

The left side integral's limits are the same: from *a* to *a*.

It's just a line at *x* = *a*, which has no area.

So the left side is 0.

The right side is *F*(*a*) + *C*.

So *C* = -*F*(*a*).

Put -*F*(*a*) into the *C*.

Then ∫_{a}^{x} *f*(*t*) *dt* = *F*(*x*) - *F*(*a*).

Put *b* into *x*.

Then you get the theorem part 2.

## Example

First, write the antiderivative of 3*x*^{2}, *x*^{3}.

Write the brackets that covers *x*^{3}.

Write the upper limit 4

at the top of the right bracket.

Write the lower limit 1

at the bottom of the right bracket.

Put 4 into *x*^{3}: 4^{3}.

This is *F*(4).

Write -.

And put 1 into *x*^{3}: 1^{3}.

This is *F*(1).

So 63 is the answer.

Keep in mind that

the result of a definite integral is a number (constant).

(if the integral limits are not variables.)

This is a big difference from an indefinite integral,

whose result is a function.