Basic Integration (AA SL)
Integration undoes differentiation: given a derivative, this skill asks you to work backwards to the original function. Every polynomial term follows the same power-rule pattern, and if you're told a point the curve passes through you can pin down the "+C" that a plain antiderivative always leaves open. It's part of the broader Integration topic.
14 questions on this sub-topic.
The power rule
Covered under IB syllabus reference SL5.5: introduction to integration as anti-differentiation of functions of the form \(f(x)=ax^n+bx^{n-1}+\dots\), for integer \(n \neq -1\), including anti-differentiation with a boundary condition. The rule below is in the formula booklet, so the challenge is applying it term by term rather than recalling it.
Power rule
\(\displaystyle\int ax^n\,dx = \dfrac{ax^{n+1}}{n+1}+C,\ n\neq-1\)
Raise the power by one, then divide by that new power. Integrate a sum of terms one term at a time.
Using a boundary condition
Given \(y\) (or \(\tfrac{dy}{dx}\)) at a specific \(x\)-value, substitute it in after integrating and solve for \(C\).
If you're integrating a second derivative, you'll do this twice - once for each constant.
Need the full syllabus wording and formula-booklet reference table? See Integration.
Worked examples
Find \(\displaystyle\int (6x^2 - 4x + 5)\,dx\).
Worked solution
\(\int\(6x^2-4x+5\)\,dx\) M1
\(=2x^3-2x^2+5x+C.\) A1
A curve satisfies \(\dfrac{d^2y}{dx^2} = 6x\). At \(x = 0\), the gradient is 2 and \(y = 1\).
Find \(y\) in terms of \(x\).
Worked solution
\(\dfrac{dy}{dx}=\int 6x\,dx=3x^2+C_1.\)M1
Gradient 2 at \(x=0\Rightarrow C_1\) \(=2.\) A1
\(y=x^3+2x+C_2.\)M1
\(y(0)=1\Rightarrow C_2\) \(=1.\) A1
Step 3 - Form as a final equation
\(y=x^3+2x+1.\) A1
A curve has \(\dfrac{dy}{dx} = 6x - 4\) and passes through \((2, 5)\).
Find \(y\).
Worked solution
\(y=\int(6x-4)\,dx=3x^2-4x+C.\) M1A1
The curve passes through \((2,5)\): \(5=3(4)-8+C=4+C\Rightarrow C=1.\) M1A1
\(y=3x^2-4x+1.\) A1
Common mistakes
- Dropping the "+C". An indefinite integral is a family of functions, not one curve - the accuracy mark is very often tied to writing \(+C\), even if the rest of the working is perfect.
- Dividing by the old power instead of the new one. \(\int x^3\,dx\) becomes \(\tfrac{x^4}{4}\), not \(\tfrac{x^4}{3}\) - the divisor is always \(n+1\), the power you just raised to.
- Muddling the two constants in a second-derivative problem. Each integration introduces its own constant, and each is fixed by a different piece of given information - match the gradient condition to \(C_1\) and the \(y\)-value condition to \(C_2\), not the other way round.
Ready to practise properly?
14 basic integration questions, marked instantly like the real exam.
Quick answers
What is the power rule for integration?
\(\displaystyle\int ax^n\,dx = \dfrac{ax^{n+1}}{n+1}+C\), for \(n \neq -1\). Raise the power by one and divide by the new power.
How do I find the constant of integration, C?
You need a boundary condition - a known \(y\)-value or gradient at a given \(x\). Substitute it into your integrated expression and solve for \(C\). See the parent Integration page for GDC guidance on checking your answer.