Fundamental Theorem of Calculus

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

The definite integral is another way to write the Riemann integral.

To find the area under the graph,
first write the sum of the area Sn (Riemann sum).

Riemann integral

As n → ∞,
Sn 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(xk) into f(x).

And change △x into dx.

So ∫ab 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)

If F(x) = [integral, from a to x, of f(t) dt], then F'(x) = f(x).

The fundamental theorem of calculus consists of two parts.

This is part 1:

If F(x) is defined as ∫ax 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)

[integral, from a to b, of f(x) dx] = F(b) - F(a)

This is part 2:

ab f(x) dx = F(b) - F(a)

This part shows how to solve a definite integral.

Proof: Part 1

Fundamental Theorem of Calculus (Part 1): Proof of the Theorem

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

Fundamental Theorem of Calculus (Part 2): Proof of the Theorem

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

Integrate (or antidifferentiate) both sides.

Then, by part 1,
the left side is ∫ax f(t) dt,
and the right side is F(x) + C.


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 ∫ax f(t) dt = F(x) - F(a).

Put b into x.

Then you get the theorem part 2.


Find the given integral. The integral, from 1 to 4, of 3x^2 dx.

First, write the antiderivative of 3x2, x3.

Write the brackets that covers x3.

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 x3: 43.
This is F(4).

Write -.
And put 1 into x3: 13.
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.