1.11.1 Fundamental Theorem of Calculus

What you'll learn

How an integral can define an accumulation function.
Why differentiating an accumulated integral recovers the original function.
How to evaluate definite integrals using antiderivatives.
How to handle variable upper limits using the chain rule.

Prerequisites

Antiderivatives

An antiderivative of a function $f (x)$ is a function whose derivative is $f (x)$ .

For example, since

ddx(x3)=3x2,\frac{d}{dx}\left(x^3\right)=3x^2,

an antiderivative of $3x^2$ is $x^3$ .

Antiderivatives are not unique: differentiating a constant gives zero, so every function of the form $x^3+C$ is also an antiderivative of $3x^2$ .

Definition

Indefinite integral

The indefinite integral represents the family of all antiderivatives of a function:

dx=F(x)+C,\int f(x)\,dx=F(x)+C,

where $F^{'} (x) = f (x)$ and $C$ is an arbitrary constant.

Definite integrals

A definite integral has a lower limit and an upper limit:

dx.\int_a^b f(x)\,dx.

Here, $a$ is the lower limit, $b$ is the upper limit, and $f (x)$ is the integrand, meaning the function being integrated.

A definite integral is a number. Geometrically, it represents signed area between the graph and the horizontal axis:

regions above the axis contribute positively;
regions below the axis contribute negatively.

Common Mistake

Integral is not always geometric area

A definite integral gives signed area. If the graph crosses the axis and the question asks for the total geometric area, split the integral at each root and make every area contribution positive.

Accumulation Functions

Suppose $f$ is continuous, and define

dt.A(x)=\int_a^x f(t)\,dt.

The value of $A (x)$ is the signed area accumulated under the graph of $f$ from the fixed point $a$ to the variable point $x$ .

The letter $t$ is a dummy variable: it is used only inside the integral. We use it so that the upper limit $x$ is not confused with the integration variable.

Accumulated area under a curve from a to x, with an additional narrow strip from x to x+h

If $x$ increases by a small amount $h$ , the extra area is approximately a narrow rectangle of width $h$ and height $f (x)$ . Therefore,

A(x+h)−A(x)≈f(x)h.A(x+h)-A(x)\approx f(x)h.

Dividing by $h$ gives

A(x+h)−A(x)h≈f(x).\frac{A(x+h)-A(x)}{h}\approx f(x).

As $h→0h\to 0$ , the approximation becomes exact. The left-hand side becomes the derivative of $A$ , so

A^{'} (x) = f (x) .

This is the first form of the Fundamental Theorem of Calculus.

Definition

Fundamental Theorem of Calculus: accumulation form

If $f$ is continuous and

dt,A(x)=\int_a^x f(t)\,dt,

then

A^{'} (x) = f (x) .

In other words,

dt)=f(x).\frac{d}{dx}\left(\int_a^x f(t)\,dt\right)=f(x).

Key Idea

Integration and differentiation are inverse processes

Integrating $f$ from a fixed lower limit to $x$ , and then differentiating the result, returns $f (x)$ .

Example

Differentiating an accumulation function

Let

dt.A(x)=\int_2^x \left(t^3+4t\right)\,dt.

Find $A^{'} (x)$ .

The integral has a fixed lower limit and upper limit $x$ , so it is in the direct accumulation form of the theorem.
Replace the dummy variable $t$ in the integrand by the upper-limit variable $x$ .
Therefore,

A'(x)=x^3+4x.

There is no need to evaluate the integral first.

Variable Upper Limits

The theorem applies directly when the upper limit is exactly $x$ . If the upper limit is a function such as $x^2$ , combine the theorem with the chain rule.

dt,A(x)=\int_a^{g(x)}f(t)\,dt,

then

A′(x)=f(g(x))g′(x).A'(x)=f\left(g(x)\right)g'(x).

You first substitute the upper limit into the integrand, then multiply by the derivative of the upper limit.

Example

Differentiating with a variable upper limit

Differentiate

dt.A(x)=\int_1^{x^2}\sqrt{1+t^3}\,dt.

The upper limit is $g(x)=x^2$ , so substitute $t=x^2$ into the integrand:

f(x2)=1+(x2)3=1+x6.f\left(x^2\right)=\sqrt{1+\left(x^2\right)^3} =\sqrt{1+x^6}.

Differentiate the upper limit:

g^{'} (x) = 2 x .

Multiply these results using the chain rule:

A′(x)=2x1+x6.A'(x)=2x\sqrt{1+x^6}.

Common Mistake

Forgetting the chain-rule factor

For an upper limit $g (x)$ , the derivative is not just $f (g (x))$ . You must multiply by $g^{'} (x)$ .

Variable lower limits

If the variable appears as the lower limit, reversing the limits introduces a minus sign:

dt.\int_x^a f(t)\,dt=-\int_a^x f(t)\,dt.

Therefore,

dt)=−f(x).\frac{d}{dx}\left(\int_x^a f(t)\,dt\right)=-f(x).

More generally,

dt)=−f(g(x))g′(x).\frac{d}{dx}\left(\int_{g(x)}^a f(t)\,dt\right) =-f\left(g(x)\right)g'(x).

Tip

Track which limit moves

A moving upper limit gives a positive contribution; a moving lower limit gives a negative contribution.

Evaluating Definite Integrals

The second form of the theorem gives the standard method for calculating a definite integral.

Definition

Fundamental Theorem of Calculus: evaluation form

If $f$ is continuous on the interval from $a$ to $b$ , and $F^{'} (x) = f (x)$ , then

dx=F(b)−F(a).\int_a^b f(x)\,dx=F(b)-F(a).

The notation

[F(x)]ab\left[F(x)\right]_a^b

means $F (b) - F (a)$ .

The theorem links the two meanings of integration:

accumulating signed area;
finding an antiderivative.

Instead of estimating the area using many narrow strips, you find an antiderivative and evaluate it at the endpoints.

Example

Evaluating a definite integral

Evaluate

dx.\int_1^3 \left(2x^2-4x+1\right)\,dx.

Find an antiderivative of the integrand:

F(x)=23x3−2x2+x.F(x)=\frac{2}{3}x^3-2x^2+x.

Apply the limits, keeping the substitution grouped:

dx=[23x3−2x2+x]13=(23(3)3−2(3)2+3)−(23(1)3−2(1)2+1).\begin{aligned} \int_1^3 \left(2x^2-4x+1\right)\,dx &=\left[\frac{2}{3}x^3-2x^2+x\right]_1^3\\ &=\left(\frac{2}{3}(3)^3-2(3)^2+3\right) -\left(\frac{2}{3}(1)^3-2(1)^2+1\right). \end{aligned}

Simplify both endpoint values:

F(3)=18−18+3=3,F(1)=23−2+1=−13.\begin{aligned} F(3)&=18-18+3=3,\\ F(1)&=\frac{2}{3}-2+1=-\frac{1}{3}. \end{aligned}

Hence,

dx=3−(−13)=103.\int_1^3 \left(2x^2-4x+1\right)\,dx =3-\left(-\frac{1}{3}\right) =\frac{10}{3}.

Common Mistake

Subtracting only part of the lower-limit value

The whole of $F (a)$ must be subtracted. Write $F(b)−(F(a))F(b)-\left(F(a)\right)$ using brackets, especially when the lower-limit value is negative.

Why there is no constant of integration

For an indefinite integral, you include $+ C$ . For a definite integral, any constant cancels:

(F(b)+C)−(F(a)+C)=F(b)−F(a).\left(F(b)+C\right)-\left(F(a)+C\right)=F(b)-F(a).

You therefore do not write $+ C$ in the final value of a definite integral.

Common Mistake

Continuity condition

The standard statement assumes that $f$ is continuous on the interval. Functions with undefined points, vertical asymptotes or other discontinuities may require methods involving improper integrals, which are not covered by simply substituting endpoints.

Exam technique

In the exam

Decide whether you are differentiating an integral or evaluating a definite integral.
For differentiation, substitute the moving limit into the integrand and multiply by its derivative; include a minus sign if it is the lower limit.
For evaluation, find an antiderivative, write the limits clearly, and calculate upper value minus lower value using brackets.
Check the sign against the graph: a function that is positive throughout the interval should give a positive integral.

Self review

Check yourself

What is $dt)\frac{d}{dx}\left(\int_4^x \cos t\,dt\right)$ ?
How would you differentiate $dt\int_0^{3x-1} e^{t^2}\,dt$ ?
Why does the constant of integration cancel when evaluating a definite integral?

What you'll learn

How an integral can define an accumulation function.
Why differentiating an accumulated integral recovers the original function.
How to evaluate definite integrals using antiderivatives.
How to handle variable upper limits using the chain rule.

Prerequisites

Antiderivatives

An antiderivative of a function $f (x)$ is a function whose derivative is $f (x)$ .

For example, since

ddx(x3)=3x2,\frac{d}{dx}\left(x^3\right)=3x^2,

an antiderivative of $3x^2$ is $x^3$ .

Antiderivatives are not unique: differentiating a constant gives zero, so every function of the form $x^3+C$ is also an antiderivative of $3x^2$ .

Definition

Indefinite integral

The indefinite integral represents the family of all antiderivatives of a function:

dx=F(x)+C,\int f(x)\,dx=F(x)+C,

where $F^{'} (x) = f (x)$ and $C$ is an arbitrary constant.

Definite integrals

A definite integral has a lower limit and an upper limit:

dx.\int_a^b f(x)\,dx.

Here, $a$ is the lower limit, $b$ is the upper limit, and $f (x)$ is the integrand, meaning the function being integrated.

A definite integral is a number. Geometrically, it represents signed area between the graph and the horizontal axis:

regions above the axis contribute positively;
regions below the axis contribute negatively.

Common Mistake

Integral is not always geometric area

A definite integral gives signed area. If the graph crosses the axis and the question asks for the total geometric area, split the integral at each root and make every area contribution positive.

Accumulation Functions

Suppose $f$ is continuous, and define

dt.A(x)=\int_a^x f(t)\,dt.

The value of $A (x)$ is the signed area accumulated under the graph of $f$ from the fixed point $a$ to the variable point $x$ .

The letter $t$ is a dummy variable: it is used only inside the integral. We use it so that the upper limit $x$ is not confused with the integration variable.

Accumulated area under a curve from a to x, with an additional narrow strip from x to x+h

If $x$ increases by a small amount $h$ , the extra area is approximately a narrow rectangle of width $h$ and height $f (x)$ . Therefore,

A(x+h)−A(x)≈f(x)h.A(x+h)-A(x)\approx f(x)h.

Dividing by $h$ gives

A(x+h)−A(x)h≈f(x).\frac{A(x+h)-A(x)}{h}\approx f(x).

As $h→0h\to 0$ , the approximation becomes exact. The left-hand side becomes the derivative of $A$ , so

A^{'} (x) = f (x) .

This is the first form of the Fundamental Theorem of Calculus.

Definition

Fundamental Theorem of Calculus: accumulation form

If $f$ is continuous and

dt,A(x)=\int_a^x f(t)\,dt,

then

A^{'} (x) = f (x) .

In other words,

dt)=f(x).\frac{d}{dx}\left(\int_a^x f(t)\,dt\right)=f(x).

Key Idea

Integration and differentiation are inverse processes

Integrating $f$ from a fixed lower limit to $x$ , and then differentiating the result, returns $f (x)$ .

Example

Differentiating an accumulation function

Let

dt.A(x)=\int_2^x \left(t^3+4t\right)\,dt.

Find $A^{'} (x)$ .

The integral has a fixed lower limit and upper limit $x$ , so it is in the direct accumulation form of the theorem.
Replace the dummy variable $t$ in the integrand by the upper-limit variable $x$ .
Therefore,

A'(x)=x^3+4x.

There is no need to evaluate the integral first.

Variable Upper Limits

The theorem applies directly when the upper limit is exactly $x$ . If the upper limit is a function such as $x^2$ , combine the theorem with the chain rule.

dt,A(x)=\int_a^{g(x)}f(t)\,dt,

then

A′(x)=f(g(x))g′(x).A'(x)=f\left(g(x)\right)g'(x).

You first substitute the upper limit into the integrand, then multiply by the derivative of the upper limit.

Example

Differentiating with a variable upper limit

Differentiate

dt.A(x)=\int_1^{x^2}\sqrt{1+t^3}\,dt.

The upper limit is $g(x)=x^2$ , so substitute $t=x^2$ into the integrand:

f(x2)=1+(x2)3=1+x6.f\left(x^2\right)=\sqrt{1+\left(x^2\right)^3} =\sqrt{1+x^6}.

Differentiate the upper limit:

g^{'} (x) = 2 x .

Multiply these results using the chain rule:

A′(x)=2x1+x6.A'(x)=2x\sqrt{1+x^6}.

Common Mistake

Forgetting the chain-rule factor

For an upper limit $g (x)$ , the derivative is not just $f (g (x))$ . You must multiply by $g^{'} (x)$ .

Variable lower limits

If the variable appears as the lower limit, reversing the limits introduces a minus sign:

dt.\int_x^a f(t)\,dt=-\int_a^x f(t)\,dt.

Therefore,

dt)=−f(x).\frac{d}{dx}\left(\int_x^a f(t)\,dt\right)=-f(x).

More generally,

dt)=−f(g(x))g′(x).\frac{d}{dx}\left(\int_{g(x)}^a f(t)\,dt\right) =-f\left(g(x)\right)g'(x).

Tip

Track which limit moves

A moving upper limit gives a positive contribution; a moving lower limit gives a negative contribution.

Evaluating Definite Integrals

The second form of the theorem gives the standard method for calculating a definite integral.

Definition

Fundamental Theorem of Calculus: evaluation form

If $f$ is continuous on the interval from $a$ to $b$ , and $F^{'} (x) = f (x)$ , then

dx=F(b)−F(a).\int_a^b f(x)\,dx=F(b)-F(a).

The notation

[F(x)]ab\left[F(x)\right]_a^b

means $F (b) - F (a)$ .

The theorem links the two meanings of integration:

accumulating signed area;
finding an antiderivative.

Instead of estimating the area using many narrow strips, you find an antiderivative and evaluate it at the endpoints.

Example

Evaluating a definite integral

Evaluate

dx.\int_1^3 \left(2x^2-4x+1\right)\,dx.

Find an antiderivative of the integrand:

F(x)=23x3−2x2+x.F(x)=\frac{2}{3}x^3-2x^2+x.

Apply the limits, keeping the substitution grouped:

dx=[23x3−2x2+x]13=(23(3)3−2(3)2+3)−(23(1)3−2(1)2+1).\begin{aligned} \int_1^3 \left(2x^2-4x+1\right)\,dx &=\left[\frac{2}{3}x^3-2x^2+x\right]_1^3\\ &=\left(\frac{2}{3}(3)^3-2(3)^2+3\right) -\left(\frac{2}{3}(1)^3-2(1)^2+1\right). \end{aligned}

Simplify both endpoint values:

F(3)=18−18+3=3,F(1)=23−2+1=−13.\begin{aligned} F(3)&=18-18+3=3,\\ F(1)&=\frac{2}{3}-2+1=-\frac{1}{3}. \end{aligned}

Hence,

dx=3−(−13)=103.\int_1^3 \left(2x^2-4x+1\right)\,dx =3-\left(-\frac{1}{3}\right) =\frac{10}{3}.

Common Mistake

Subtracting only part of the lower-limit value

The whole of $F (a)$ must be subtracted. Write $F(b)−(F(a))F(b)-\left(F(a)\right)$ using brackets, especially when the lower-limit value is negative.

Why there is no constant of integration

For an indefinite integral, you include $+ C$ . For a definite integral, any constant cancels:

(F(b)+C)−(F(a)+C)=F(b)−F(a).\left(F(b)+C\right)-\left(F(a)+C\right)=F(b)-F(a).

You therefore do not write $+ C$ in the final value of a definite integral.

Common Mistake

Continuity condition

Exam technique

In the exam

Decide whether you are differentiating an integral or evaluating a definite integral.
For differentiation, substitute the moving limit into the integrand and multiply by its derivative; include a minus sign if it is the lower limit.
For evaluation, find an antiderivative, write the limits clearly, and calculate upper value minus lower value using brackets.
Check the sign against the graph: a function that is positive throughout the interval should give a positive integral.

Self review

Check yourself

What is $dt)\frac{d}{dx}\left(\int_4^x \cos t\,dt\right)$ ?
How would you differentiate $dt\int_0^{3x-1} e^{t^2}\,dt$ ?
Why does the constant of integration cancel when evaluating a definite integral?

What you'll learn

Prerequisites

Antiderivatives

Definite integrals

Accumulation Functions

Variable Upper Limits

Variable lower limits

Evaluating Definite Integrals

Why there is no constant of integration

Review 1.11.1 Fundamental Theorem of Calculus by teaching Genie

1.11.1 Fundamental Theorem of Calculus

What you'll learn

Prerequisites

Antiderivatives

Definite integrals

Accumulation Functions

Variable Upper Limits

Variable lower limits

Evaluating Definite Integrals

Why there is no constant of integration

Review 1.11.1 Fundamental Theorem of Calculus by teaching Genie

1.11.1 Fundamental Theorem of Calculus

What you'll learn

Prerequisites

Antiderivatives

Definite integrals

Accumulation Functions

Variable Upper Limits

Variable lower limits

Evaluating Definite Integrals

Why there is no constant of integration

Review 1.11.1 Fundamental Theorem of Calculus by teaching Genie

1.11.1 Fundamental Theorem of Calculus Revision Guide

1.11.1 Fundamental Theorem of Calculus

What you'll learn

Prerequisites

Antiderivatives

Definite integrals

Accumulation Functions

Variable Upper Limits

Variable lower limits

Evaluating Definite Integrals

Why there is no constant of integration

Review 1.11.1 Fundamental Theorem of Calculus by teaching Genie

1.11.1 Fundamental Theorem of Calculus Revision Guide