How do you integrate a function with a linear inner, like (2x+1)^4?

Use the rule integral of (ax+b)^r equals (ax+b) to the power r+1, divided by a times (r+1), plus C. For (2x+1) to the power 4, a equals 2 and r equals 4, so the answer is (2x+1) to the power 5 divided by 10, plus C. The key step is dividing by the inner coefficient a as well as by r+1.

How do you integrate a function with a linear inner, like (2x+1)^4?

Use the rule integral of (ax+b)^r equals (ax+b) to the power r+1, divided by a times (r+1), plus C. For (2x+1) to the power 4, a equals 2 and r equals 4, so the answer is (2x+1) to the power 5 divided by 10, plus C. The key step is dividing by the inner coefficient a as well as by r+1.

Year 12 Methods (Unit 3 & 4) Integration

Further integration techniques

Q: What is the fundamental theorem of calculus?

The fundamental theorem says the definite integral from a to b of f(x) dx equals F(b) minus F(a), where F is any anti-derivative of f. You anti-differentiate, substitute the upper limit, substitute the lower limit, and subtract. No constant of integration is needed because it cancels.

Q: How do you integrate a product or a fraction?

There is no product or quotient rule for integration, so simplify first. Expand a product such as (x+2)(x-2) into x squared minus 4, split a fraction term by term such as (x squared plus 1) over x into x plus 1 over x, or write a root as a power such as root x equals x to the power one half. Then integrate each standard term.

Q: How do you use a given derivative to find an integral?

Anti-differentiation reverses differentiation. If you are told that the derivative of F(x) is f(x), then the integral of f(x) dx is F(x) plus C. For example, since the derivative of x times e to the x equals e to the x times (x+1), the integral of e to the x times (x+1) dx is x times e to the x plus C.

Q: How do you find the area under a curve?

If f(x) is positive between x equals a and x equals b, the area enclosed by the curve and the x-axis equals the definite integral from a to b of f(x) dx, evaluated as F(b) minus F(a). For example the area under y equals root x from x equals 1 to x equals 4 is 14 over 3, about 4.67 square units.

Q: How do you find f(x) from f'(x) and a point?

Anti-differentiate f'(x) to get f(x) with an unknown constant C, then substitute the given point to solve for C. For f'(x) equals 6x squared minus 4 with f(2) equals 5, anti-differentiating gives 2x cubed minus 4x plus C; using f(2) equals 5 gives C equals negative 3, so f(x) equals 2x cubed minus 4x minus 3.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), further integration brings the standard anti-derivative forms together and evaluates them. You integrate the standard families (keeping the \(\tfrac{1}{k}\) or \(\tfrac{1}{a}\) factor for a linear inner), simplify a product or fraction first, reverse a given derivative, and use the fundamental theorem \(\int_a^b f(x)\,dx=F(b)-F(a)\) to find an area or a total change.

Anti-differentiation is the reverse of differentiation: an anti-derivative \(F\) of \(f\) satisfies \(F'(x)=f(x)\). The indefinite integral \(\int f(x)\,dx=F(x)+C\) collects all anti-derivatives through the constant of integration \(C\).

A definite integral \(\int_a^b f(x)\,dx\) is a number. The fundamental theorem of calculus evaluates it:

\(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a).\)

When \(f(x)>0\) on \([a,b]\), this number is the area enclosed by the curve \(y=f(x)\) and the \(x\)-axis between \(x=a\) and \(x=b\). Because there is no product or quotient rule for integration, an awkward integrand is first simplified into standard forms — expanding a product, splitting a fraction, or writing a root as a power.

Key idea. Anti-differentiate to get \(F\); an indefinite integral needs \(+C\). A definite integral is \(F(b)-F(a)\) (no \(+C\)), and equals the area under the curve when \(f>0\).

Area \(=\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\)

\(\displaystyle\int_1^4 \sqrt{x}\,dx=\dfrac{14}{3}\approx 4.67\) square units

The standard anti-derivatives (\(C\) is the constant of integration):

\[\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\ (n\neq-1),\quad \int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+C,\quad \int\dfrac{1}{x}\,dx=\ln|x|+C\]

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

\[\int\sin(kx)\,dx=-\dfrac{1}{k}\cos(kx)+C,\quad \int\cos(kx)\,dx=\dfrac{1}{k}\sin(kx)+C\]

Linear inner \(ax+b\) — keep the \(\dfrac{1}{a}\):

\[\int(ax+b)^{r}\,dx=\dfrac{(ax+b)^{r+1}}{a\,(r+1)}+C,\quad \int\dfrac{1}{ax+b}\,dx=\dfrac{1}{a}\ln|ax+b|+C\]

\int (a x + b)^{r} d x = \frac{(a x + b)^{r + 1}}{a (r + 1)} + C

The fundamental theorem of calculus:

\[\int_a^b f(x)\,dx=F(b)-F(a)\]

\int_{a}^{b} f (x) d x = F (b) - F (a)

Simplify first. No product or quotient rule: expand \((x+2)(x-2)=x^{2}-4\), split \(\dfrac{x^{2}+1}{x}=x+\dfrac{1}{x}\), or write \(\sqrt{x}=x^{1/2}\), then integrate each standard term.

How to work a further-integration problem

Simplify the integrand. Expand a product, split a fraction term by term, or write a root as a power so every term is a standard form.
Integrate each standard term. Apply the power rule, or the \(e^{kx}\), \(\sin(kx)\), \(\cos(kx)\), \(\dfrac{1}{ax+b}\) forms — keeping the \(\dfrac{1}{k}\) or \(\dfrac{1}{a}\) factor for a linear inner. Add \(+C\) for an indefinite integral.
Reverse a given derivative. If \(\dfrac{d}{dx}\big(F(x)\big)=f(x)\) is supplied, then \(\int f(x)\,dx=F(x)+C\) at once.
Evaluate a definite integral. Use \(\int_a^b f(x)\,dx=F(b)-F(a)\): substitute the upper limit minus the lower limit (no \(+C\)).

Boundary point. To find \(f(x)\) from \(f'(x)\), anti-differentiate to get \(f(x)+C\), then substitute a known point \(f(a)=b\) to solve for \(C\) — the one particular anti-derivative the problem asks for.

Example 1 — Linear inner

Find \(\displaystyle\int(2x+1)^{4}\,dx\).

Solution

Raise the power and divide by \(a(r+1)=2\times5\).

\(\displaystyle\int(2x+1)^{4}\,dx\)	\(=\)	\(\dfrac{(2x+1)^{5}}{2\times5}+C\)
	\(=\)	\(\dfrac{1}{10}(2x+1)^{5}+C\)

\frac{1}{10} (2 x + 1)^{5} + C

Example 2 — Simplify first

Evaluate \(\displaystyle\int_{1}^{2}\dfrac{x^{2}+1}{x}\,dx\).

Solution

Split the fraction, then use \(F(b)-F(a)\).

\(\dfrac{x^{2}+1}{x}\)	\(=\)	\(x+\dfrac{1}{x}\)
\(\displaystyle\int_{1}^{2}\!\Big(x+\tfrac{1}{x}\Big)dx\)	\(=\)	\(\Big[\tfrac{x^{2}}{2}+\ln x\Big]_{1}^{2}\)
	\(=\)	\(\tfrac{3}{2}+\ln 2\approx 2.19\)

\frac{3}{2} + \ln 2

Example 3 — Given derivative

Given \(\dfrac{d}{dx}\!\left(xe^{x}\right)=e^{x}(x+1)\), find \(\displaystyle\int e^{x}(x+1)\,dx\).

Solution

Integration reverses the given derivative.

\(\displaystyle\int e^{x}(x+1)\,dx\)

\(=\)

\(xe^{x}+C\)

Check: \(\dfrac{d}{dx}(xe^{x})=e^{x}+xe^{x}=e^{x}(x+1).\)

x e^{x} + C

Example 4 — Boundary point

\(f'(x)=6x^{2}-4\) and \(f(2)=5\). Find \(f(x)\).

Solution

Anti-differentiate, then use the point to find \(C\).

\(f(x)\)	\(=\)	\(2x^{3}-4x+C\)
\(16-8+C\)	\(=\)	\(5\)
\(C\)	\(=\)	\(-3\)
\(f(x)\)	\(=\)	\(2x^{3}-4x-3\)

f (x) = 2 x^{3} - 4 x - 3

Common pitfalls

Keep the \(\tfrac{1}{a}\) or \(\tfrac{1}{k}\) factor. For a linear inner, \(\int e^{3x}\,dx=\tfrac{1}{3}e^{3x}+C\) and \(\int(2x+1)^{4}\,dx=\tfrac{1}{10}(2x+1)^{5}+C\); dropping the factor is the most common error.

Simplify before integrating. There is no product or quotient rule, so expand a product or split a fraction into standard terms first — you cannot integrate factor by factor.

Mind the \(+C\) and the order of limits. An indefinite integral needs \(+C\); a definite integral does not. Always subtract \(F(a)\) from \(F(b)\) — swapping the limits flips the sign.

Frequently asked questions

How do you integrate (2x+1)^4?

Raise the power to \(5\) and divide by \(a(r+1)=2\times5=10\): \(\int(2x+1)^{4}\,dx=\tfrac{1}{10}(2x+1)^{5}+C\).

What is the fundamental theorem of calculus?

\(\int_a^b f(x)\,dx=F(b)-F(a)\), where \(F\) is any anti-derivative of \(f\). Anti-differentiate, then substitute the upper limit minus the lower limit.

How do you integrate a product or a fraction?

Simplify first. Expand \((x+2)(x-2)=x^{2}-4\), split \(\dfrac{x^{2}+1}{x}=x+\dfrac{1}{x}\), or write \(\sqrt{x}=x^{1/2}\), then integrate each standard term.

How do you use a given derivative to find an integral?

If \(\dfrac{d}{dx}(F(x))=f(x)\) is given, then \(\int f(x)\,dx=F(x)+C\). For example, since \(\dfrac{d}{dx}(xe^{x})=e^{x}(x+1)\), \(\int e^{x}(x+1)\,dx=xe^{x}+C\).

How do you find the area under a curve?

If \(f(x)>0\) on \([a,b]\), the area is \(\int_a^b f(x)\,dx=F(b)-F(a)\). For example \(\int_1^4 \sqrt{x}\,dx=\tfrac{14}{3}\approx 4.67\) square units.

How do you find f(x) from f'(x) and a point?

Anti-differentiate to get \(f(x)+C\), then substitute the point to solve for \(C\). For \(f'(x)=6x^{2}-4\), \(f(2)=5\): \(f(x)=2x^{3}-4x+C\) and \(C=-3\), so \(f(x)=2x^{3}-4x-3\).

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