How do you integrate a product like x squared times (x minus 3)?

There is no product rule for integration, so expand the product into a sum of powers first, then integrate term by term. So x squared times (x minus 3) becomes x cubed minus 3x squared, whose integral is x to the fourth over 4 minus x cubed plus C. Quotients are simplified to a sum of powers in the same way before integrating.

How do you find the particular anti-derivative through a given point?

First anti-differentiate to get the general form ending in plus C. Then substitute the coordinates of the given point and solve for C. For example if f dash equals 3x squared minus 4x and the curve passes through (1,2), then f equals x cubed minus 2x squared plus C, and 2 equals 1 minus 2 plus C gives C equals 3, so f equals x cubed minus 2x squared plus 3.

How do you integrate a product like x squared times (x minus 3)?

There is no product rule for integration, so expand the product into a sum of powers first, then integrate term by term. So x squared times (x minus 3) becomes x cubed minus 3x squared, whose integral is x to the fourth over 4 minus x cubed plus C. Quotients are simplified to a sum of powers in the same way before integrating.

How do you find the particular anti-derivative through a given point?

First anti-differentiate to get the general form ending in plus C. Then substitute the coordinates of the given point and solve for C. For example if f dash equals 3x squared minus 4x and the curve passes through (1,2), then f equals x cubed minus 2x squared plus C, and 2 equals 1 minus 2 plus C gives C equals 3, so f equals x cubed minus 2x squared plus 3.

Year 12 Methods (Unit 3 & 4) Anti-differentiation

Anti-differentiation of polynomial functions

Q: What is anti-differentiation?

Anti-differentiation is the reverse of differentiation: given a derivative or rate of change, you find a function it could have come from. The result is called an indefinite integral, written with the integral sign, and it always includes a constant of integration plus C because many functions share the same derivative.

Q: How do you anti-differentiate a power of x?

Use the reverse power rule: the integral of x to the power n is x to the power n plus 1, divided by n plus 1, plus C. In words, raise the index by one and divide by the new index. For example the integral of x squared is x cubed over 3 plus C, and the integral of a constant k is k times x plus C.

Q: Why do you always add plus C?

Because the derivative of any constant is zero, adding any constant to an anti-derivative gives another function with the same derivative. So an indefinite integral stands for a whole family of functions, and the constant of integration plus C represents that freedom. You leave it out only when a boundary condition is given to find its value.

Q: How do you integrate a polynomial with several terms?

Use linearity: integrate each term separately and add the results, and a constant multiple can come out the front. For example the integral of 3x squared minus 4x plus 5 is x cubed minus 2x squared plus 5x plus C, where the constant term 5 integrates to 5x. Only one plus C is needed for the whole expression.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), anti-differentiation is the reverse of differentiation. For a polynomial you use the reverse power rule \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\), integrate term by term by linearity, and always include the constant of integration \(+C\) — then a boundary point fixes \(C\) to give the particular anti-derivative, ready for definite integrals in Unit 4.

Anti-differentiation (or finding the indefinite integral) reverses differentiation: given \(f'(x)\), you find a function \(f(x)\) whose derivative is \(f'(x)\), written \(\displaystyle\int f'(x)\,dx\).

For a power of \(x\) the key tool is the reverse power rule (for a non-negative integer \(n\)):

\(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C.\)

Raise the index by \(1\), divide by the new index, and add the constant of integration \(+C\). Because the derivative of any constant is \(0\), infinitely many functions share the same derivative, so the \(+C\) is always required. By linearity, a whole polynomial integrates term by term; a product or quotient is expanded or simplified to a sum of powers first.

Key idea. \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\): raise the index, divide by the new index, add \(+C\). A boundary point \((a,b)\) fixes \(C\).

The family \(y=x^{2}+C\): same gradient \(2x\), shifted by \(C\)

The boundary point \((1,2)\) selects one member: \(C=3\)

The reverse power rule (for a non-negative integer \(n\)):

\[\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C,\qquad \int k\,dx=kx+C\]

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

Linearity — integrate term by term and pull out constant multiples:

\[\int \big(f(x)+g(x)\big)\,dx=\int f(x)\,dx+\int g(x)\,dx,\qquad \int k\,f(x)\,dx=k\int f(x)\,dx\]

\int k f (x) d x = k \int f (x) d x

Particular anti-derivative — use a boundary point \((a,b)\):

\[f(x)=\int f'(x)\,dx+C,\qquad b=f(a) \ \Rightarrow\ C\]

Expand or simplify first. There is no product or quotient rule for integration. Rewrite a product such as \(x^{2}(x-3)\) or a quotient such as \(\dfrac{2x^{4}-6x^{2}}{2x^{2}}\) as a sum of powers, then integrate term by term.

How to anti-differentiate a polynomial

Rewrite as a sum of powers. Expand any product and simplify any quotient so every term is a power \(k x^{n}\).
Apply the reverse power rule to each term. Raise the index by \(1\), divide by the new index; a constant \(k\) integrates to \(kx\).
Add one \(+C\). A single constant of integration covers the whole expression.
Fix \(C\) if a point is given. Substitute the boundary point \((a,b)\) and solve for \(C\) to get the particular anti-derivative.

Modelling. A rate of change integrates back to the quantity. Anti-differentiate the polynomial rate, then use the known starting value (e.g. the amount at \(t=0\)) to find \(C\) before evaluating.

Example 1 — Term by term

Find \(\displaystyle\int (3x^{2}-4x+5)\,dx\).

Solution

Integrate each term; the constant \(5\) gives \(5x\).

\(\displaystyle\int (3x^{2}-4x+5)\,dx\)	\(=\)	\(3\cdot\dfrac{x^{3}}{3}-4\cdot\dfrac{x^{2}}{2}+5x+C\)
	\(=\)	\(x^{3}-2x^{2}+5x+C\)

x^{3} - 2 x^{2} + 5 x + C

Example 2 — Expand first

Find \(\displaystyle\int x^{2}(x-3)\,dx\).

Solution

Expand to a sum of powers, then integrate.

\(\displaystyle\int x^{2}(x-3)\,dx\)	\(=\)	\(\displaystyle\int (x^{3}-3x^{2})\,dx\)
	\(=\)	\(\dfrac{x^{4}}{4}-x^{3}+C\)

\frac{x^{4}}{4} - x^{3} + C

Example 3 — Particular anti-derivative

\(f'(x)=3x^{2}-4x\) and the curve passes through \((1,2)\). Find \(f(x)\).

Solution

Integrate, then substitute \((1,2)\) to find \(C\).

\(f(x)\)	\(=\)	\(x^{3}-2x^{2}+C\)
\(2\)	\(=\)	\(1-2+C\Rightarrow C=3\)
\(f(x)\)	\(=\)	\(x^{3}-2x^{2}+3\)

f (x) = x^{3} - 2 x^{2} + 3

Example 4 — Rate model

Water flows in at \(\dfrac{dV}{dt}=3t^{2}+4\) litres/min, with \(V(0)=5\) litres. Find \(V(t)\).

Solution

Integrate the rate, then use \(V(0)=5\) to find \(C\).

\(V\)	\(=\)	\(t^{3}+4t+C\)
\(5\)	\(=\)	\(0+0+C\Rightarrow C=5\)
\(V(t)\)	\(=\)	\(t^{3}+4t+5\)

V (t) = t^{3} + 4 t + 5

Common pitfalls

Raise the index, do not divide by the old one. \(\displaystyle\int x^{4}\,dx=\dfrac{x^{5}}{5}+C\), not \(\dfrac{x^{4}}{4}\) and not \(\dfrac{x^{4}}{5}\); the index must go up by one before you divide.

Never forget \(+C\). An indefinite integral stands for a whole family of functions; leaving off the constant of integration loses that. Only drop it once a boundary condition has fixed its value.

Use the whole point when solving for \(C\). Substitute both coordinates and keep the constant term; at \((0,b)\) every power of \(0\) vanishes, so \(C=b\).

Frequently asked questions

What is anti-differentiation?

It is the reverse of differentiation: from a derivative you recover a function whose derivative it is. The result, the indefinite integral, always carries a constant of integration \(+C\).

How do you anti-differentiate a power of x?

Use \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\): raise the index by one and divide by the new index. So \(\displaystyle\int x^{2}\,dx=\dfrac{x^{3}}{3}+C\).

Why do you always add plus C?

Because the derivative of any constant is \(0\), so infinitely many functions share the same derivative. The \(+C\) captures that whole family; a boundary point pins down one member.

How do you integrate a polynomial with several terms?

By linearity, integrate each term separately with one \(+C\). For example \(\displaystyle\int (3x^{2}-4x+5)\,dx=x^{3}-2x^{2}+5x+C\), where \(5\) becomes \(5x\).

How do you integrate a product like x^2(x-3)?

There is no product rule for integration, so expand first: \(x^{2}(x-3)=x^{3}-3x^{2}\), giving \(\dfrac{x^{4}}{4}-x^{3}+C\). Quotients are simplified the same way.

How do you find the particular anti-derivative through a point?

Integrate to the general form ending in \(+C\), then substitute the point and solve for \(C\). E.g. \(f'=3x^{2}-4x\) through \((1,2)\) gives \(C=3\), so \(f=x^{3}-2x^{2}+3\).

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