Integrating Powers of x (DP IB Applications & Interpretation (AI): SL): Revision Note

Integrating Powers of x

How do I integrate powers of x? 

• Powers of $x$ are integrated according to the following formula:

  • If $\mathrm{f}\left(x\right)=x^n$ then $\int \mathrm{f}\left(x\right)\mathrm{d}x=\frac{x^{n+1}}{n+1}+c$ 

    • where $n\in \mathrm{\mathbb{Z}},n\ne -1$ 

    • and $c$ is the constant of integration

• This is given in the formula booklet

• If the power of is $x$ multiplied by a constant then the integral is also multiplied by that constant

  • If $\mathrm{f}\left(x\right)=a{x}^n$ then $\int \mathrm{f}\left(x\right)\mathrm{d}x=\frac{a{x}^{n+1}}{n+1}+c$ 

    • where $n\in \mathrm{\mathbb{Z}},n\ne -1$

    •  $a$ is a constant

    • and $c$ is the constant of integration

• Don’t forget the special case:

  • $\int a\mathrm{d}x=ax+c$

    • e.g.  $\int 4\mathrm{d}x=4x+c$

  • This allows constant terms to be integrated

• Functions involving fractions with denominators in terms of $x$ will need to be rewritten as negative powers of $x$ first

  • e.g.  If $\mathrm{f}\left(x\right)=\frac{4}{x^2}$ then rewrite as $\mathrm{f}\left(x\right)=4x^{-2}$ and integrate

How do I integrate sums and differences of powers of x?

• To integrate a sum or difference of power of$x$ terms, just integrate term by term

  • e.g.  If $\mathrm{f}\left(x\right)=8x^3-2x+4$ then $\int \mathrm{f}\left(x\right)\mathrm{d}x=\frac{8x^{3+1}}{3+1}-\frac{2x^{1+1}}{1+1}+4x+c=2x^4-x^2+4x+c$

• Products and quotients cannot be integrated in this way so would need expanding/simplifying first

  • e.g. If $\mathrm{f}\left(x\right)=8x^2(2x-3)$ then $\int \mathrm{f}\left(x\right)\mathrm{d}x=\int (16x^3-24x^2)\mathrm{d}x=\frac{16x^4}{4}-\frac{24x^3}{3}+c=4x^4-8x^3+c$