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iGCSE (2021 Edition)

12.04 Integration of exponential functions


Recall that we previously established the following results when differentiating exponential functions:

Differentiating exponential functions
$\frac{d}{dx}e^x$ddxex $=$= $e^x$ex
$\frac{d}{dx}e^{ax+b}$ddxeax+b $=$= $ae^{ax+b}$aeax+b

Reversing these we get the following rules for integrating exponential functions:

Integration of exponential functions

$\int e^xdx=e^x+C$exdx=ex+C, for some constant $C$C

$\int e^{ax+b}dx=\frac{1}{a}e^{ax+b}+C$eax+bdx=1aeax+b+C, for some constant $C$C


Worked examples

Example 1

Determine $\int e^{2-7x}dx$e27xdx.

Think: To integrate we need to divide by $a$a from the term $ax+b$ax+b; here $a=-7$a=7. The power of $e$e remains unchanged.


$\int e^{2-7x}dx=\frac{-1}{7}e^{2-7x}+C$e27xdx=17e27x+C, where $C$C is a constant.

Example 2

Determine $\int\frac{e^{2x}-3}{e^x}dx$e2x3exdx.

Think: We can split this fraction and rewrite the terms as powers of $e$e. Then we can apply our rule term by term.


$\frac{e^{2x}-3}{e^x}$e2x3ex $=$= $\frac{e^{2x}}{e^x}-\frac{3}{e^x}$e2xex3ex

Split the fraction

  $=$= $e^{2x-x}-3e^{-x}$e2xx3ex

Use index laws to rewrite

  $=$= $e^x-3e^{-x}$ex3ex




$\int\frac{e^{2x}-3}{e^x}dx$e2x3exdx $=$= $\int e^x-3e^{-x}dx$ex3exdx
  $=$= $e^x+3e^{-x}+C$ex+3ex+C, where $C$C is a constant
Example 3

If $f'\left(x\right)=1+8e^{-2x}$f(x)=1+8e2x and $f\left(0\right)=2$f(0)=2, find $f\left(x\right)$f(x).

Think: We can first find the indefinite integral, and then use the given point $\left(0,2\right)$(0,2) to find the value of the constant of integration.


$\int1+8e^{-2x}dx$1+8e2xdx $=$= $x-4e^{-2x}+C$x4e2x+C


Using the point $\left(0,2\right)$(0,2), find $C$C:

$f\left(0\right)$f(0) $=$= $2$2
$\therefore\ 0-4e^0+C$ 04e0+C $=$= $2$2
$-4+C$4+C $=$= $2$2
$C$C $=$= $6$6

Thus, $f\left(x\right)=x-4e^{-2x}+6$f(x)=x4e2x+6.


Practice questions

Question 1

Determine $\int4e^{1-2x}dx$4e12xdx.

You may use $C$C as a constant.

Question 2

A curve has gradient function $\frac{dy}{dx}=e^{3x}$dydx=e3x. Find the equation of the curve if it passes through $\left(0,-1\right)$(0,1).

Question 3

Given $\frac{dy}{dx}=ke^x+2$dydx=kex+2, for some constant $k$k, and that when $x=0$x=0 we have $y=-1$y=1 and $\frac{dy}{dx}=5$dydx=5:

  1. Determine the value of $k$k.

  2. Find $\int\left(ke^x+2\right)dx$(kex+2)dx using the value of $k$k from part (a). Use $C$C as the unknown constant.

  3. Find $y$y in terms of $x$x.



Integrate functions of the form e^(ax + b).

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