Mathematics 9709 · Pure Mathematics 2

Integration of exponential functions

Integration is differentiation run backwards. It undoes a derivative.
When you differentiate $e^{ax+b}$ , the chain rule drops an extra $a$ out front. To undo that, you divide by a.
That one idea (reverse the derivative, then put the $\tfrac1a$ back in) gives every formula in this chapter: exponentials, logs and trig.

$\int e^{ax+b}\,dx=\tfrac1a\,e^{ax+b}+c$ . On its own $e^x$ integrates to itself. The only new part is the $\tfrac1a$ . It cancels the extra $a$ from the chain rule.

The rule, and where the 1/a comes from

Differentiating $e^{ax+b}$ gives $a\,e^{ax+b}$ , so to reverse it you divide that extra $a$ back out again:

$\int e^{ax+b}\,dx=\tfrac1a\,e^{ax+b}+c\qquad(a\ne0)$

A constant out front stays where it is: $\int k\,e^{ax+b}\,dx=\tfrac{k}{a}\,e^{ax+b}+c$ .

Check by differentiating your answer. It has to give back what you started with. A dropped $\tfrac1a$ or a sign slip shows up straight away. This check works for every answer in this chapter.

Worked examples

Worked example

Find $\int 3e^{2x}\,dx$

Read off $k=3$ (the constant) and $a=2$ (the coefficient of $x$ ).
Reverse $e^{2x}$ to itself, then divide by $a=2$ : $\int 3e^{2x}\,dx=\tfrac32 e^{2x}+c$ .
Check: differentiate $\tfrac32 e^{2x}$ back to $\tfrac32\cdot 2\,e^{2x}=3e^{2x}$ . ✓

Answer

\tfrac32 e^{2x}+c

Worked example

Find $\int e^{1-x}\,dx$

The inside is $-x+1$ , so the number in front of $x$ is $a=-1$ (the trap is reading it as $+1$ ), and dividing by $-1$ just flips the sign:

$\int e^{1-x}\,dx=\tfrac{1}{-1}e^{1-x}+c=-e^{1-x}+c$

Answer

-e^{1-x}+c

Worked example

Evaluate $\int_1^2 9e^{3x-1}\,dx$ (modelled on a P3 exponential integral)

A definite integral: integrate first, then put the limits in and do top minus bottom.

Integrate (the constant $9$ divided by $a=3$ gives $3$ ): $\int 9e^{3x-1}\,dx=3e^{3x-1}$ , so we evaluate $\big[\,3e^{3x-1}\,\big]_1^2$ .
Substitute the top limit $x=2$ : $3e^{6-1}=3e^{5}$ . Then the bottom limit $x=1$ : $3e^{3-1}=3e^{2}$ .
Top minus bottom: $3e^{5}-3e^{2}$ . Leave it exact — no decimals unless asked.

Answer

3e^{5}-3e^{2}

Why dividing by a undoes the chain rule

You don't memorise the 1/a — it has to be there

Say the answer were just $e^{ax+b}$ . Differentiate it. The chain rule gives $a\,e^{ax+b}$ , with an extra $a$ you never wanted. To get rid of that $a$ ahead of time, start with $\tfrac1a\,e^{ax+b}$ . Its derivative is $\tfrac1a\cdot a\,e^{ax+b}=e^{ax+b}$ exactly. So the $\tfrac1a$ is not a random rule. It is whatever number it takes to make differentiation give back what you started with. That is why the same $\tfrac1a$ shows up again for $\tfrac1{ax+b}$ , $\sin(ax+b)$ and the rest.

Where the marks go

Common mistake

You can't skip the

\tfrac1a

\int e^{2x}\,dx=\tfrac12 e^{2x}+c

, never

e^{2x}+c

. And an indefinite integral always needs the

+c

: drop it and you drop a mark.

Common mistake

Read

a

with its sign. In

e^{1-x}

the number in front of

x

-1

, so the answer is

-e^{1-x}+c

, not

+e^{1-x}+c

Now you try

Evaluate $\int_0^1\big(e^{2x}+2e^{-x}\big)\,dx$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Integrate one term at a time — each $e^{ax}$ gets its own $\tfrac1a$ : $\big[\tfrac12 e^{2x}-2e^{-x}\big]_0^1=\left(\tfrac12 e^{2}-\tfrac2e\right)-\left(\tfrac12-2\right)$ :

Answer

\tfrac12 e^{2}+\tfrac32-\tfrac{2}{e}

PreviousParametric Differentiation NextIntegration of 1/(ax + b)

Integration of exponential functions

Integration is differentiation run backwards. It undoes a derivative.
When you differentiate $e^{ax+b}$ , the chain rule drops an extra $a$ out front. To undo that, you divide by a.
That one idea (reverse the derivative, then put the $\tfrac1a$ back in) gives every formula in this chapter: exponentials, logs and trig.

$\int e^{ax+b}\,dx=\tfrac1a\,e^{ax+b}+c$ . On its own $e^x$ integrates to itself. The only new part is the $\tfrac1a$ . It cancels the extra $a$ from the chain rule.

The rule, and where the 1/a comes from

Differentiating $e^{ax+b}$ gives $a\,e^{ax+b}$ , so to reverse it you divide that extra $a$ back out again:

$\int e^{ax+b}\,dx=\tfrac1a\,e^{ax+b}+c\qquad(a\ne0)$

A constant out front stays where it is: $\int k\,e^{ax+b}\,dx=\tfrac{k}{a}\,e^{ax+b}+c$ .

Check by differentiating your answer. It has to give back what you started with. A dropped $\tfrac1a$ or a sign slip shows up straight away. This check works for every answer in this chapter.

Worked examples

Worked example

Find $\int 3e^{2x}\,dx$

Read off $k=3$ (the constant) and $a=2$ (the coefficient of $x$ ).
Reverse $e^{2x}$ to itself, then divide by $a=2$ : $\int 3e^{2x}\,dx=\tfrac32 e^{2x}+c$ .
Check: differentiate $\tfrac32 e^{2x}$ back to $\tfrac32\cdot 2\,e^{2x}=3e^{2x}$ . ✓

Answer

\tfrac32 e^{2x}+c

Worked example

Find $\int e^{1-x}\,dx$

The inside is $-x+1$ , so the number in front of $x$ is $a=-1$ (the trap is reading it as $+1$ ), and dividing by $-1$ just flips the sign:

$\int e^{1-x}\,dx=\tfrac{1}{-1}e^{1-x}+c=-e^{1-x}+c$

Answer

-e^{1-x}+c

Worked example

Evaluate $\int_1^2 9e^{3x-1}\,dx$ (modelled on a P3 exponential integral)

A definite integral: integrate first, then put the limits in and do top minus bottom.

Integrate (the constant $9$ divided by $a=3$ gives $3$ ): $\int 9e^{3x-1}\,dx=3e^{3x-1}$ , so we evaluate $\big[\,3e^{3x-1}\,\big]_1^2$ .
Substitute the top limit $x=2$ : $3e^{6-1}=3e^{5}$ . Then the bottom limit $x=1$ : $3e^{3-1}=3e^{2}$ .
Top minus bottom: $3e^{5}-3e^{2}$ . Leave it exact — no decimals unless asked.

Answer

3e^{5}-3e^{2}

Why dividing by a undoes the chain rule

You don't memorise the 1/a — it has to be there

Where the marks go

Common mistake

You can't skip the

\tfrac1a

\int e^{2x}\,dx=\tfrac12 e^{2x}+c

, never

e^{2x}+c

. And an indefinite integral always needs the

+c

: drop it and you drop a mark.

Common mistake

Read

a

with its sign. In

e^{1-x}

the number in front of

x

-1

, so the answer is

-e^{1-x}+c

, not

+e^{1-x}+c

Now you try

Evaluate $\int_0^1\big(e^{2x}+2e^{-x}\big)\,dx$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Integrate one term at a time — each $e^{ax}$ gets its own $\tfrac1a$ : $\big[\tfrac12 e^{2x}-2e^{-x}\big]_0^1=\left(\tfrac12 e^{2}-\tfrac2e\right)-\left(\tfrac12-2\right)$ :

Answer

\tfrac12 e^{2}+\tfrac32-\tfrac{2}{e}