IB Maths AA 5.4 Notes

Anti-differentiation

Integration is the reverse of differentiation. If differentiation gives the rate of change of a function, then integration rebuilds the original function from its derivative. This begins with anti-differentiation:

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

If $F'(x)=f(x)$, then $F(x)$ is an antiderivative of $f(x)$. The constant $C$ is the constant of integration.

The constant is needed because differentiating any constant gives $0$. For example, since $\frac{d}{dx}(x^2)=2x$ and $\frac{d}{dx}(x^2+5)=2x$, we have $\int 2x\,dx=x^2+C$.

$\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$

For $n\in\mathbb{Z}$ and $n\ne-1$, $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$.

More generally:

$\int ax^n\,dx=a\frac{x^{n+1}}{n+1}+C$

A few examples include:

• $\int 3x^2\,dx=x^3+C$
• $\int 4x^3\,dx=x^4+C$
• $\int (4x^3+2x)\,dx=x^4+x^2+C$
• $\int (6x^4-5x+1)\,dx=\frac{6}{5}x^5-\frac{5}{2}x^2+x+C$

Sometimes the constant of integration can be found using extra information.

Integrate $\frac{dy}{dx}=3x^2+x$ if $y=10$ when $x=1$.

First integrate: $y=x^3+\frac{1}{2}x^2+C$.

Substitute $x=1$ and $y=10$: $10=1+\frac{1}{2}+C$.

So $10=1.5+C$, giving $C=8.5=\frac{17}{2}$.

Hence $y=x^3+\frac{1}{2}x^2+8.5$.