1.8.1 Fundamental theorem of calculus (A-level only)

What you'll learn

How an integral with a variable limit defines an accumulation function.
Why differentiating an accumulation function recovers the original function.
How to evaluate definite integrals using antiderivatives.
How to handle variable limits, reversed limits and signed areas correctly.

Prerequisites

Antiderivatives

An antiderivative of a function $f$ is a function $F$ whose derivative is $f$ :

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

For example, because

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

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

Antiderivatives are not unique. Every function of the form $x^3+C$ , where $C$ is a constant, has derivative $3x^2$ .

Definition

Indefinite integral

The notation

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

represents the family of all antiderivatives of $f$ , where $F^{'} (x) = f (x)$ and $C$ is an arbitrary constant.

Example

Finding an antiderivative

Find an antiderivative of $f(x)=6x^2-4x+3$ .

Integrate each term using the reverse power rule:
$dx=xn+1n+1+C,n≠−1.\int x^n\,dx=\frac{x^{n+1}}{n+1}+C, \qquad n\neq -1.$
Applying this term by term gives
$dx=6(x33)−4(x22)+3x+C=2x3−2x2+3x+C.\begin{aligned} \int (6x^2-4x+3)\,dx &=6\left(\frac{x^3}{3}\right)-4\left(\frac{x^2}{2}\right)+3x+C\\ &=2x^3-2x^2+3x+C. \end{aligned}$
A quick differentiation check gives
$ddx(2x3−2x2+3x+C)=6x2−4x+3,\frac{d}{dx}\left(2x^3-2x^2+3x+C\right)=6x^2-4x+3,$
confirming that the antiderivative is correct.

Definite integrals and accumulation

A definite integral has fixed or variable limits. Geometrically, it gives the signed area between a curve and the horizontal axis.

For a continuous function $f$ ,

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

counts area above the horizontal axis positively and area below it negatively.

An accumulation function records how much signed area has accumulated as its upper limit moves. Fix a starting point $a$ and define

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

The letter $t$ is a dummy variable: it is used only inside the integral. The variable $x$ controls the upper limit and is therefore the input of $A$ .

The shaded region below represents $A (x)$ . Moving the upper boundary from $x$ to $x + h$ adds a thin strip whose area is approximately $f (x) h$ .

A curve y = f(t) with the accumulated area from a to x shaded and a thin extra strip from x to x+h

Key Idea

Rate of accumulation

If $x$ increases by a small amount $h$ , then the accumulated area changes by approximately

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

Dividing by $h$ suggests that the rate at which area accumulates is $A^{'} (x) = f (x)$ .

The first part of the theorem

Definition

Fundamental theorem of calculus: differentiation form

If $f$ is continuous, then

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

In words: differentiating an integral with variable upper limit recovers the original integrand, evaluated at that upper limit.

This result makes the inverse relationship between integration and differentiation precise. Integration accumulates values of a function; differentiation measures the rate of that accumulation.

Example

Differentiating an accumulation function

Differentiate

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

The lower limit is constant and the upper limit is exactly $x$ , so the differentiation form of the theorem applies directly.
Replace the dummy variable $t$ in the integrand by the upper limit $x$ :
$A'(x)=x^3+4x.$
There is no need to evaluate the integral first: the theorem gives the derivative immediately.

Common Mistake

Keeping the dummy variable

After differentiating $dt\int_a^x f(t)\,dt$ , write the answer as $f (x)$ , not $f (t)$ . The derivative is a function of $x$ .

When the upper limit is a function

Suppose the upper limit is $g (x)$ rather than simply $x$ :

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

The fundamental theorem gives the derivative with respect to the upper limit, but the chain rule must then account for the rate at which that limit changes:

A′(x)=f(g(x))g′(x).A'(x)=f\bigl(g(x)\bigr)g'(x).

Example

Differentiating with a variable upper limit

Differentiate

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

Evaluate the integrand at the upper limit $t=x^2$ :
$1+(x2)3=1+x6.\sqrt{1+(x^2)^3}=\sqrt{1+x^6}.$
Differentiate the upper limit:
$ddx(x2)=2x.\frac{d}{dx}\left(x^2\right)=2x.$
Multiply the two results using the chain rule:
$A′(x)=2x1+x6.A'(x)=2x\sqrt{1+x^6}.$

Tip

Upper-limit pattern

For $dt)\dfrac{d}{dx}\left(\int_a^{g(x)}f(t)\,dt\right)$ , use the pattern “integrand at the limit, multiplied by the derivative of the limit”.

A variable lower limit

Reversing the limits changes the sign:

dt.\int_{g(x)}^a f(t)\,dt=-\int_a^{g(x)}f(t)\,dt.

Therefore,

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

Example

Differentiating with a variable lower limit

Differentiate

dt.B(x)=\int_{x^3}^{4}\cos t\,dt.

Since the variable expression is the lower limit, reverse the limits and introduce a minus sign:
$dt.B(x)=-\int_4^{x^3}\cos t\,dt.$
Evaluate the integrand at $t=x^3$ and differentiate the limit $x^3$ :
$cos⁡(x3)⋅3x2.\cos(x^3)\cdot 3x^2.$
Include the minus sign caused by reversing the limits:
$B'(x)=-3x^2\cos(x^3).$

The second part of the theorem

The theorem also explains how antiderivatives can be used to calculate definite integrals exactly.

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).

This is often written as

dx=[F(x)]ab.\int_a^b f(x)\,dx=\left[F(x)\right]_a^b.

The notation $[F(x)]ab\left[F(x)\right]_a^b$ means “substitute the upper limit, then subtract the result from substituting the lower limit”.

The constant of integration is unnecessary in a definite integral because it cancels:

(F(b)+C)−(F(a)+C)=F(b)−F(a).\bigl(F(b)+C\bigr)-\bigl(F(a)+C\bigr)=F(b)-F(a).

Example

Evaluating a definite integral

Evaluate

dx.\int_1^3 (2x^2-4x)\,dx.

Find an antiderivative:
$dx=23x3−2x2.\int (2x^2-4x)\,dx=\frac{2}{3}x^3-2x^2.$
Substitute the upper and lower limits:
$dx=[23x3−2x2]13.\int_1^3 (2x^2-4x)\,dx =\left[\frac{2}{3}x^3-2x^2\right]_1^3.$
Evaluate the upper limit first:
$23(33)−2(32)=18−18=0.\frac{2}{3}(3^3)-2(3^2)=18-18=0.$
Evaluate the lower limit and subtract it:
$0−(23−2)=0−(−43)=43.0-\left(\frac{2}{3}-2\right) =0-\left(-\frac{4}{3}\right) =\frac{4}{3}.$

Common Mistake

Forgetting the lower-limit brackets

Always calculate $F (b) - F (a)$ with brackets around the entire lower-limit value. Otherwise, one or more terms may be subtracted incorrectly.

Integral versus geometric area

A definite integral gives signed area, which is not always the same as total geometric area.

If $f (x) < 0$ , then

dx<0\int_a^b f(x)\,dx<0

over an interval where the graph lies below the horizontal axis. Geometric area must be positive, so you must split at any points where the curve crosses the axis and make the below-axis contribution positive.

Example

Finding total area across the axis

Find the total area between $y = x - 1$ , the horizontal axis, and the lines $x = 0$ and $x = 3$ .

Solve $x - 1 = 0$ to find where the graph crosses the axis:
$x = 1.$
The graph is below the axis from 0 to 1 and above it from 1 to 3.
Find the signed integral on the first interval:
$dx=[x22−x]01=−12.\int_0^1(x-1)\,dx =\left[\frac{x^2}{2}-x\right]_0^1 =-\frac{1}{2}.$
Its geometric area is therefore $12\frac{1}{2}$ .
Find the area on the second interval:
$dx=[x22−x]13=2.\int_1^3(x-1)\,dx =\left[\frac{x^2}{2}-x\right]_1^3 =2.$
Add the positive areas:
$area=12+2=52.\text{Total area}=\frac{1}{2}+2=\frac{5}{2}.$

Common Mistake

Continuity matters

The standard form of the theorem assumes that the integrand is continuous over the relevant interval. Functions with discontinuities or infinite values may require separate treatment as improper integrals.

Useful consequences

The evaluation form immediately gives these properties:

dx=0,\int_a^a f(x)\,dx=0,

because no interval is traversed, and

dx,\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx,

because reversing the direction reverses the sign.

It also shows why differentiation and integration are inverse processes:

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

while

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

Exam technique

In the exam

Decide whether you are differentiating an integral or evaluating one: the two forms of the theorem perform different jobs.
For a variable limit $g (x)$ , substitute it into the integrand and multiply by $g^{'} (x)$ ; include a minus sign if it is the lower limit.
For a definite integral, find an antiderivative and use upper minus lower, keeping the lower-limit substitution in brackets.
If the question asks for geometric area, identify axis crossings and split the integral before making each contribution positive.

Self review

Check yourself

Can you explain why $dt)=f(x)\dfrac{d}{dx}\left(\int_a^x f(t)\,dt\right)=f(x)$ using the area of a thin strip?
How would you differentiate $dt\displaystyle \int_2^{\sin x}(1+t^2)\,dt$ ?
Why might the value of a definite integral differ from the total area between a graph and the horizontal axis?

What you'll learn

How an integral with a variable limit defines an accumulation function.
Why differentiating an accumulation function recovers the original function.
How to evaluate definite integrals using antiderivatives.
How to handle variable limits, reversed limits and signed areas correctly.

Prerequisites

Antiderivatives

An antiderivative of a function $f$ is a function $F$ whose derivative is $f$ :

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

For example, because

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

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

Antiderivatives are not unique. Every function of the form $x^3+C$ , where $C$ is a constant, has derivative $3x^2$ .

Definition

Indefinite integral

The notation

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

represents the family of all antiderivatives of $f$ , where $F^{'} (x) = f (x)$ and $C$ is an arbitrary constant.

Example

Finding an antiderivative

Find an antiderivative of $f(x)=6x^2-4x+3$ .

Integrate each term using the reverse power rule:
$dx=xn+1n+1+C,n≠−1.\int x^n\,dx=\frac{x^{n+1}}{n+1}+C, \qquad n\neq -1.$
Applying this term by term gives
$dx=6(x33)−4(x22)+3x+C=2x3−2x2+3x+C.\begin{aligned} \int (6x^2-4x+3)\,dx &=6\left(\frac{x^3}{3}\right)-4\left(\frac{x^2}{2}\right)+3x+C\\ &=2x^3-2x^2+3x+C. \end{aligned}$
A quick differentiation check gives
$ddx(2x3−2x2+3x+C)=6x2−4x+3,\frac{d}{dx}\left(2x^3-2x^2+3x+C\right)=6x^2-4x+3,$
confirming that the antiderivative is correct.

Definite integrals and accumulation

A definite integral has fixed or variable limits. Geometrically, it gives the signed area between a curve and the horizontal axis.

For a continuous function $f$ ,

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

counts area above the horizontal axis positively and area below it negatively.

An accumulation function records how much signed area has accumulated as its upper limit moves. Fix a starting point $a$ and define

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

The letter $t$ is a dummy variable: it is used only inside the integral. The variable $x$ controls the upper limit and is therefore the input of $A$ .

The shaded region below represents $A (x)$ . Moving the upper boundary from $x$ to $x + h$ adds a thin strip whose area is approximately $f (x) h$ .

A curve y = f(t) with the accumulated area from a to x shaded and a thin extra strip from x to x+h

Key Idea

Rate of accumulation

If $x$ increases by a small amount $h$ , then the accumulated area changes by approximately

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

Dividing by $h$ suggests that the rate at which area accumulates is $A^{'} (x) = f (x)$ .

The first part of the theorem

Definition

Fundamental theorem of calculus: differentiation form

If $f$ is continuous, then

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

In words: differentiating an integral with variable upper limit recovers the original integrand, evaluated at that upper limit.

This result makes the inverse relationship between integration and differentiation precise. Integration accumulates values of a function; differentiation measures the rate of that accumulation.

Example

Differentiating an accumulation function

Differentiate

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

The lower limit is constant and the upper limit is exactly $x$ , so the differentiation form of the theorem applies directly.
Replace the dummy variable $t$ in the integrand by the upper limit $x$ :
$A'(x)=x^3+4x.$
There is no need to evaluate the integral first: the theorem gives the derivative immediately.

Common Mistake

Keeping the dummy variable

After differentiating $dt\int_a^x f(t)\,dt$ , write the answer as $f (x)$ , not $f (t)$ . The derivative is a function of $x$ .

When the upper limit is a function

Suppose the upper limit is $g (x)$ rather than simply $x$ :

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

The fundamental theorem gives the derivative with respect to the upper limit, but the chain rule must then account for the rate at which that limit changes:

A′(x)=f(g(x))g′(x).A'(x)=f\bigl(g(x)\bigr)g'(x).

Example

Differentiating with a variable upper limit

Differentiate

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

Evaluate the integrand at the upper limit $t=x^2$ :
$1+(x2)3=1+x6.\sqrt{1+(x^2)^3}=\sqrt{1+x^6}.$
Differentiate the upper limit:
$ddx(x2)=2x.\frac{d}{dx}\left(x^2\right)=2x.$
Multiply the two results using the chain rule:
$A′(x)=2x1+x6.A'(x)=2x\sqrt{1+x^6}.$

Tip

Upper-limit pattern

For $dt)\dfrac{d}{dx}\left(\int_a^{g(x)}f(t)\,dt\right)$ , use the pattern “integrand at the limit, multiplied by the derivative of the limit”.

A variable lower limit

Reversing the limits changes the sign:

dt.\int_{g(x)}^a f(t)\,dt=-\int_a^{g(x)}f(t)\,dt.

Therefore,

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

Example

Differentiating with a variable lower limit

Differentiate

dt.B(x)=\int_{x^3}^{4}\cos t\,dt.

Since the variable expression is the lower limit, reverse the limits and introduce a minus sign:
$dt.B(x)=-\int_4^{x^3}\cos t\,dt.$
Evaluate the integrand at $t=x^3$ and differentiate the limit $x^3$ :
$cos⁡(x3)⋅3x2.\cos(x^3)\cdot 3x^2.$
Include the minus sign caused by reversing the limits:
$B'(x)=-3x^2\cos(x^3).$

The second part of the theorem

The theorem also explains how antiderivatives can be used to calculate definite integrals exactly.

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).

This is often written as

dx=[F(x)]ab.\int_a^b f(x)\,dx=\left[F(x)\right]_a^b.

The notation $[F(x)]ab\left[F(x)\right]_a^b$ means “substitute the upper limit, then subtract the result from substituting the lower limit”.

The constant of integration is unnecessary in a definite integral because it cancels:

(F(b)+C)−(F(a)+C)=F(b)−F(a).\bigl(F(b)+C\bigr)-\bigl(F(a)+C\bigr)=F(b)-F(a).

Example

Evaluating a definite integral

Evaluate

dx.\int_1^3 (2x^2-4x)\,dx.

Find an antiderivative:
$dx=23x3−2x2.\int (2x^2-4x)\,dx=\frac{2}{3}x^3-2x^2.$
Substitute the upper and lower limits:
$dx=[23x3−2x2]13.\int_1^3 (2x^2-4x)\,dx =\left[\frac{2}{3}x^3-2x^2\right]_1^3.$
Evaluate the upper limit first:
$23(33)−2(32)=18−18=0.\frac{2}{3}(3^3)-2(3^2)=18-18=0.$
Evaluate the lower limit and subtract it:
$0−(23−2)=0−(−43)=43.0-\left(\frac{2}{3}-2\right) =0-\left(-\frac{4}{3}\right) =\frac{4}{3}.$

Common Mistake

Forgetting the lower-limit brackets

Always calculate $F (b) - F (a)$ with brackets around the entire lower-limit value. Otherwise, one or more terms may be subtracted incorrectly.

Integral versus geometric area

A definite integral gives signed area, which is not always the same as total geometric area.

If $f (x) < 0$ , then

dx<0\int_a^b f(x)\,dx<0

Example

Finding total area across the axis

Find the total area between $y = x - 1$ , the horizontal axis, and the lines $x = 0$ and $x = 3$ .

Solve $x - 1 = 0$ to find where the graph crosses the axis:
$x = 1.$
The graph is below the axis from 0 to 1 and above it from 1 to 3.
Find the signed integral on the first interval:
$dx=[x22−x]01=−12.\int_0^1(x-1)\,dx =\left[\frac{x^2}{2}-x\right]_0^1 =-\frac{1}{2}.$
Its geometric area is therefore $12\frac{1}{2}$ .
Find the area on the second interval:
$dx=[x22−x]13=2.\int_1^3(x-1)\,dx =\left[\frac{x^2}{2}-x\right]_1^3 =2.$
Add the positive areas:
$area=12+2=52.\text{Total area}=\frac{1}{2}+2=\frac{5}{2}.$

Common Mistake

Continuity matters

Useful consequences

The evaluation form immediately gives these properties:

dx=0,\int_a^a f(x)\,dx=0,

because no interval is traversed, and

dx,\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx,

because reversing the direction reverses the sign.

It also shows why differentiation and integration are inverse processes:

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

while

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

Exam technique

In the exam

Decide whether you are differentiating an integral or evaluating one: the two forms of the theorem perform different jobs.
For a variable limit $g (x)$ , substitute it into the integrand and multiply by $g^{'} (x)$ ; include a minus sign if it is the lower limit.
For a definite integral, find an antiderivative and use upper minus lower, keeping the lower-limit substitution in brackets.
If the question asks for geometric area, identify axis crossings and split the integral before making each contribution positive.

Self review

Check yourself

Can you explain why $dt)=f(x)\dfrac{d}{dx}\left(\int_a^x f(t)\,dt\right)=f(x)$ using the area of a thin strip?
How would you differentiate $dt\displaystyle \int_2^{\sin x}(1+t^2)\,dt$ ?
Why might the value of a definite integral differ from the total area between a graph and the horizontal axis?

What you'll learn

Prerequisites

Antiderivatives

Definite integrals and accumulation

The first part of the theorem

When the upper limit is a function

A variable lower limit

The second part of the theorem

Integral versus geometric area

Useful consequences

Review 1.8.1 Fundamental theorem of calculus (A-level only) by teaching Genie

1.8.1 Fundamental theorem of calculus (A-level only)

What you'll learn

Prerequisites

Antiderivatives

Definite integrals and accumulation

The first part of the theorem

When the upper limit is a function

A variable lower limit

The second part of the theorem

Integral versus geometric area

Useful consequences

Review 1.8.1 Fundamental theorem of calculus (A-level only) by teaching Genie

1.8.1 Fundamental theorem of calculus (A-level only)

What you'll learn

Prerequisites

Antiderivatives

Definite integrals and accumulation

The first part of the theorem

When the upper limit is a function

A variable lower limit

The second part of the theorem

Integral versus geometric area

Useful consequences

Review 1.8.1 Fundamental theorem of calculus (A-level only) by teaching Genie

1.8.1 Fundamental theorem of calculus (A-level only) Revision Guide

1.8.1 Fundamental theorem of calculus (A-level only)

What you'll learn

Prerequisites

Antiderivatives

Definite integrals and accumulation

The first part of the theorem

When the upper limit is a function

A variable lower limit

The second part of the theorem

Integral versus geometric area

Useful consequences

Review 1.8.1 Fundamental theorem of calculus (A-level only) by teaching Genie

1.8.1 Fundamental theorem of calculus (A-level only) Revision Guide