5.5 Introduction to integration (IB Maths: Analysis and Approaches SL): Revision notes

Section 1

Integration as anti-differentiation

Integration reverses differentiation. If $F'(x)=f(x)$ , then $F$ is an anti-derivative of $f$ and we write $\int f(x)\,\mathrm{d}x=F(x)+C.$ For any integer $n\neq-1$ : $\int ax^{n}\,\mathrm{d}x=\frac{a}{n+1}x^{n+1}+C.$ Raise the power by one, then divide by the new power. Integrate term by term.

Example: $\int\left(6x^{2}-4x+3\right)\mathrm{d}x=2x^{3}-2x^{2}+3x+C$ .

Key termsanti-derivativeindefinite integral

Common mistake

Dividing by the old power: $\int x^{3}\,\mathrm{d}x=\frac{x^{4}}{4}+C$ , not $\frac{x^{4}}{3}+C$ .

Exam tip

Check by differentiating your answer: you should get back the integrand.

Section 2

Negative powers and the constant of integration

Rewrite fractions as powers first: $\int\frac{2}{x^{2}}\,\mathrm{d}x=\int2x^{-2}\,\mathrm{d}x=\frac{2x^{-1}}{-1}+C=-\frac{2}{x}+C.$ The rule fails for $n=-1$ (you would divide by 0), so $\int x^{-1}\,\mathrm{d}x$ is not covered here.

The constant of integration $C$ is needed because constants differentiate to zero: $x^{2}+1$ , $x^{2}-7$ and $x^{2}$ all have derivative $2x$ . An indefinite integral without $+C$ loses a mark.

Key termsconstant of integration

Common mistake

Raising $-2$ to $-3$ when integrating $x^{-2}$ . For integration the power goes up: $-2\to-1$ .

Section 3

Boundary conditions

A boundary condition (a known point on the curve, or a starting value) fixes $C$ . Integrate first, then substitute.

Example: $\frac{\mathrm{d}y}{\mathrm{d}x}=6x^{2}-4x+3$ and $y=5$ when $x=1$ : $y=2x^{3}-2x^{2}+3x+C$ , $5=2-2+3+C$ , so $C=2$ and $y=2x^{3}-2x^{2}+3x+2$ .

In context: if water flows in at $r(t)$ litres per minute and the tank starts with 50 litres, $V(t)=\int r(t)\,\mathrm{d}t$ with $V(0)=50$ .

Key termsboundary condition

Common mistake

Setting $C$ equal to the $y$ -value of the point. You must substitute both coordinates into the integrated equation.

Section 4

Definite integrals

A definite integral has limits: $\int_{a}^{b}f(x)\,\mathrm{d}x=\big[F(x)\big]_{a}^{b}=F(b)-F(a).$ The constant $C$ cancels, so it is left out. It gives a number, not a function.

If $f$ is a rate of change, the definite integral is the total change between $a$ and $b$ : water flowing in at $12t-t^{2}$ litres per minute gives $\int_{0}^{12}(12t-t^{2})\,\mathrm{d}t=288$ litres over 12 minutes.

On Paper 2, evaluate definite integrals with your GDC, but write down the integral you are evaluating.

Key termsdefinite integrallimits

Exam tip

Brackets matter: $F(b)-F(a)$ with negative lower limits, e.g. $(24-8)-(-24+8)=32$ .

Section 5

Area under a curve

If $f(x)>0$ for $a<x<b$ , the area of the region enclosed by $y=f(x)$ , the $x$ -axis and the lines $x=a$ , $x=b$ is $A=\int_{a}^{b}f(x)\,\mathrm{d}x.$ Always write the integral expression first: it earns marks on its own.

If the region is enclosed by the curve and the $x$ -axis only, the limits are the $x$ -intercepts. Example: $f(x)=12-3x^{2}$ meets the axis at $x=\pm2$ , so $A=\int_{-2}^{2}(12-3x^{2})\,\mathrm{d}x=32$ .

Key termsarea under a curve

Common mistake

Using $y$ -values (such as the $y$ -intercept) as limits. The limits of $\int\ldots\mathrm{d}x$ are $x$ -values.

Must know

$\int ax^{n}\,\mathrm{d}x=\frac{a}{n+1}x^{n+1}+C$ , $n\neq-1$ ; always include $+C$ .
Rewrite $\frac{a}{x^{n}}$ as $ax^{-n}$ before integrating.
Use a boundary condition to find $C$ .
$\int_{a}^{b}f(x)\,\mathrm{d}x=F(b)-F(a)$ ; use technology on Paper 2 but write the integral.
Area for $f(x)>0$ : write $\int_{a}^{b}f(x)\,\mathrm{d}x$ first, limits are $x$ -values.

That's the notes covered.

Carry on to the next subtopic.