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New Zealand
Level 8 - NCEA Level 3

Primitives, indefinite, definite integrals


The Primitive

Having established the result $\frac{\mathrm{d}}{\mathrm{d}x}e^{ax+b}=ae^{ax+b}$ddxeax+b=aeax+b, then it follows that the primitive of the function $f(x)=e^{ax+b}$f(x)=eax+b is given by $F(x)=\frac{1}{a}e^{ax+b}+C$F(x)=1aeax+b+C


Note carefully that this observation was dependent on the fact that the exponent $ax+b$ax+b is linear. It is not generally true that the primitive of say $g\left(x\right)=e^{u\left(x\right)}$g(x)=eu(x) is given by $G\left(x\right)=\frac{1}{u'\left(x\right)}\times e^{u\left(x\right)}+C$G(x)=1u(x)×eu(x)+C. This is a common misconception caused by students over-generalising simpler results.  



Here is a table listing a few results using this rule:

$f(x)$f(x) $F(x)$F(x)
$e^{-4x}$e4x $-\frac{1}{4}e^{-4x}+C$14e4x+C
$6e^{2-3x}$6e23x $-2e^{2-3x}+C$2e23x+C
$\frac{4}{e^{2x+1}}=4e^{-\left(2x+1\right)}$4e2x+1=4e(2x+1) $\frac{-2}{e^{2x+1}}+C$2e2x+1+C
$e^{\frac{1}{2}x}+\sqrt{x}$e12x+x $2e^{\frac{1}{2}x}+\frac{2}{3}x\sqrt{x}+C$2e12x+23xx+C
$\frac{e^x+e^{-x}}{2}$ex+ex2 $\frac{e^x-e^{-x}}{2}+C$exex2+C


Indefinite integrals

We can interpret the rule for primitives in terms of indefinite integrals so that:

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

Hence, for example:

 $\int e^{4x}dx=\frac{1}{4}e^{4x}+C$e4xdx=14e4x+C and $\int3e^{2-7x}dx=-\frac{3}{7}e^{2-7x}+C$3e27xdx=37e27x+C.


Anti-differentiation strategies

There are however other instances where the idea of anti-differentiation can be used to established more complex results.

As a simple example, suppose we determine correctly that $\frac{\mathrm{d}}{\mathrm{d}x}e^{x^2}=2xe^{x^2}$ddxex2=2xex2.

Then we also know that:

$\int xe^{x^2}dx=\frac{1}{2}\int2xe^{x^2}dx=\frac{1}{2}e^{x^2}+C$xex2dx=122xex2dx=12ex2+C

We could develop from this a general strategy by noting that if a certain integral is in the form $\int f'(x)e^{f(x)}dx$f(x)ef(x)dx, then we can immediately conclude that:

$\int f'(x)e^{f(x)}dx=e^{f(x)}+C$f(x)ef(x)dx=ef(x)+C

Slightly more difficult situations arise where integrals can be determined indirectly. In certain circumstances, a known differentiation result can be treated as an equation that can be manipulated to find an integration result. Carefully consider this next example: 

Suppose we first establish the following result using the product rule:


Then we can determine the integral $\int xe^xdx$xexdx with the following strategy:

$\frac{\mathrm{d}}{\mathrm{d}x}xe^x$ddxxex $=$= $e^x(x+1)$ex(x+1)
$\frac{\mathrm{d}}{\mathrm{d}x}xe^x$ddxxex $=$= $xe^x+e^x$xex+ex
$\frac{\mathrm{d}}{\mathrm{d}x}xe^x-e^x$ddxxexex $=$= $xe^x$xex
$\int\left(\frac{\mathrm{d}}{\mathrm{d}x}xe^x\right)dx-\int e^xdx$(ddxxex)dxexdx $=$= $\int xe^xdx$xexdx
$xe^x-e^x$xexex $=$= $\int xe^xdx$xexdx
$\therefore\int xe^xdx$xexdx $=$= $xe^x-e^x$xexex


Definite Integrals

Again, the same principles apply for definite integrals;

Example 1


$\int_1^312e^{3x+1}dx$3112e3x+1dx $=$= $\left[4e^{3x+1}\right]_1^3$[4e3x+1]31
  $=$= $4e^{10}-4e^4$4e104e4
  $=$= $4e^4\left(e^6-1\right)$4e4(e61)
Example 2


$\int_1^3xe^{1-x^2}dx$31xe1x2dx $=$= $-\frac{1}{2}\int_1^3-2xe^{1-x^2}dx$12312xe1x2dx
  $=$= $-\frac{1}{2}\left[e^{1-x^2}\right]_1^3$12[e1x2]31
  $=$= $-\frac{1}{2}\left(e^{-8}-e^0\right)$12(e8e0)
  $=$= $\frac{e^0-e^{-8}}{2}$e0e82

More Worked Examples


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

You may use $C$C as a constant.


Find the exact value of $\int_0^3\left(2e^{5x}+e^{-7x}\right)dx$30(2e5x+e7x)dx.


Consider the following.

  1. Given that $y=e^{3x}\left(x-\frac{1}{3}\right)$y=e3x(x13), determine $y'$y.

    You may use the substitutions $u=e^{3x}$u=e3x and $v=\left(x-\frac{1}{3}\right)$v=(x13) in your working.

  2. Hence find the exact value of $\int_6^9xe^{3x}dx$96xe3xdx.



Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods


Apply integration methods in solving problems

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