How do you integrate a function with a linear inner part like e to the kx or (ax+b) to the r?

Integrate as if the inner function were x, then divide by the coefficient of x. So the integral of e to the kx is one over k times e to the kx plus C; the integral of (ax+b) to the r is (ax+b) to the r plus one, over a times r plus one, plus C; and the integral of one over ax plus b is one over a times the natural log of the absolute value of ax plus b, plus C. The plus b inside does not change the factor because the inner function is still linear with slope a.

Why does one over x integrate to a logarithm?

The power rule for integrating x to the n divides by n plus one, but at n equals minus one that would divide by zero, so the rule fails. Instead the integral of one over x is the natural logarithm of the absolute value of x, plus C, because the derivative of the natural log of x is one over x. The absolute value covers negative x as well.

How do you integrate a sum of several different forms?

Use linearity. The integral of a sum is the sum of the integrals, and a constant multiple can be taken outside the integral sign. So you integrate each term with its own standard form and keep a single constant of integration plus C. For example the integral of six x squared plus e to the two x minus cosine three x is two x cubed plus one half e to the two x minus one third sine three x plus C.

How do you integrate a function with a linear inner part like e to the kx or (ax+b) to the r?

Integrate as if the inner function were x, then divide by the coefficient of x. So the integral of e to the kx is one over k times e to the kx plus C; the integral of (ax+b) to the r is (ax+b) to the r plus one, over a times r plus one, plus C; and the integral of one over ax plus b is one over a times the natural log of the absolute value of ax plus b, plus C. The plus b inside does not change the factor because the inner function is still linear with slope a.

Why does one over x integrate to a logarithm?

The power rule for integrating x to the n divides by n plus one, but at n equals minus one that would divide by zero, so the rule fails. Instead the integral of one over x is the natural logarithm of the absolute value of x, plus C, because the derivative of the natural log of x is one over x. The absolute value covers negative x as well.

How do you integrate a sum of several different forms?

Use linearity. The integral of a sum is the sum of the integrals, and a constant multiple can be taken outside the integral sign. So you integrate each term with its own standard form and keep a single constant of integration plus C. For example the integral of six x squared plus e to the two x minus cosine three x is two x cubed plus one half e to the two x minus one third sine three x plus C.

Year 12 Methods (Unit 3 & 4) Integration

Integration of more families of functions

Q: What are the standard integrals in Methods?

The integral of x to the n is x to the n plus one over n plus one, plus C, for n not equal to minus one. The integral of e to the x is e to the x plus C. The integral of one over x is the natural log of the absolute value of x, plus C. The integral of sine x is minus cosine x plus C, and the integral of cosine x is sine x plus C. Every indefinite integral carries the constant of integration plus C.

Q: How do you evaluate a definite integral?

Use the fundamental theorem of calculus. Anti-differentiate to get F of x, with no plus C, then substitute the upper limit b and the lower limit a and subtract, giving F of b minus F of a. Use exact values so the arithmetic stays exact. For example the integral of one over x from one to e equals the natural log of x from one to e, which is log e minus log one, that is one minus zero, giving one.

Q: How is a definite integral related to area?

If a function f is greater than or equal to zero on the interval from a to b, then the area between the curve and the x-axis over that interval equals the definite integral of f from a to b. The same idea gives a total change from a rate, such as displacement from a velocity or the volume delivered from a flow rate.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), a small set of standard integrals covers every family of function: \(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\) (for \(n\neq-1\)), \(\displaystyle\int e^{x}\,dx=e^{x}+C\), \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\), \(\displaystyle\int \sin x\,dx=-\cos x+C\) and \(\displaystyle\int \cos x\,dx=\sin x+C\). For a linear inner \(ax+b\) you divide by \(a\); by linearity you integrate several forms in one go; and the fundamental theorem of calculus evaluates a definite integral \(\displaystyle\int_{a}^{b} f(x)\,dx=F(b)-F(a)\) as the area under a curve.

Integration (finding the indefinite integral) reverses differentiation: given \(f(x)\) you find \(F(x)\) with \(F'(x)=f(x)\), written \(\displaystyle\int f(x)\,dx=F(x)+C\).

The standard forms for the Methods families are:

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

\(\displaystyle\int \sin x\,dx=-\cos x+C,\qquad \int \cos x\,dx=\sin x+C\)

When the inner function is linear, \(f(ax+b)\), integrate as if the inner were \(x\) and then divide by the coefficient \(a\) of \(x\). This one rule generates \(e^{kx}\), \((ax+b)^{r}\) and \(\dfrac{1}{ax+b}\). Every indefinite integral carries the constant of integration \(+C\); all angles are in radians.

Key idea. Match the term to its standard form, divide by the coefficient of \(x\) for a linear inner, add \(+C\). For a definite integral use \(\displaystyle\int_{a}^{b} f(x)\,dx=F(b)-F(a)\).

\(\displaystyle\int_{1}^{e}\dfrac{1}{x}\,dx=1\): the reciprocal family read as an area

Area \(=\displaystyle\int_{0}^{1}2e^{x}\,dx=2(e-1)\approx 3.44\)

Standard forms (add \(+C\); angles in radians):

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

\[\int \sin x\,dx=-\cos x+C,\qquad \int \cos x\,dx=\sin x+C\]

\int \frac{1}{x} d x = \ln | x | + C

Linear inner function — divide by the coefficient \(a\) of \(x\):

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

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

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

Definite integral (fundamental theorem of calculus) — and area when \(f(x)\ge 0\):

\[\int_{a}^{b} f(x)\,dx=\Big[F(x)\Big]_{a}^{b}=F(b)-F(a)\]

Check by differentiating. The derivative of your anti-derivative must return the original integrand — a fast way to catch a missing \(\tfrac{1}{a}\) factor or a sign slip.

How to integrate across the families

Match each term to its standard form: a power \(x^{n}\), \(e^{x}\), \(\dfrac{1}{x}\), \(\sin x\) or \(\cos x\).
Handle a linear inner \(ax+b\). Integrate as if the inner were \(x\), then divide by the coefficient \(a\) (this covers \(e^{kx}\), \((ax+b)^{r}\) and \(\dfrac{1}{ax+b}\)).
Use linearity. Integrate a sum term by term and take constant multiples out the front, keeping a single \(+C\).
For a particular anti-derivative, substitute the given point into \(F(x)+C\) and solve for \(C\).
For a definite integral, anti-differentiate (no \(+C\)), then compute \(F(b)-F(a)\) with exact values; if \(f\ge 0\) this is the area under the curve.

Total change. Integrating a rate over an interval gives the total change — displacement from a velocity, or the volume delivered from a flow rate.

Example 1 — Several forms at once

Find \(\displaystyle\int \left(6x^{2}+e^{2x}-\cos 3x\right)\,dx\).

Solution

Integrate term by term; divide \(e^{2x}\) by \(2\) and \(\cos 3x\) by \(3\).

\(\displaystyle\int \left(6x^{2}+e^{2x}-\cos 3x\right)\,dx\)

\(=\)

\(2x^{3}+\dfrac{1}{2}e^{2x}-\dfrac{1}{3}\sin 3x+C\)

2 x^{3} + \frac{1}{2} e^{2 x} - \frac{1}{3} \sin 3 x + C

Example 2 — A linear inner power

Find \(\displaystyle\int (2x+1)^{3}\,dx\).

Solution

Raise the index by one, then divide by \(a(r+1)=2\times 4\).

\(\displaystyle\int (2x+1)^{3}\,dx\)	\(=\)	\(\dfrac{(2x+1)^{4}}{2\times 4}+C\)
	\(=\)	\(\dfrac{(2x+1)^{4}}{8}+C\)

\frac{{(2 x + 1)}^{4}}{8} + C

Example 3 — Definite integral of the reciprocal

Evaluate \(\displaystyle\int_{1}^{4}\dfrac{6}{2x+1}\,dx\).

Solution

\(\int\dfrac{1}{2x+1}\,dx=\dfrac{1}{2}\ln(2x+1)\), so the factor is \(6\times\dfrac{1}{2}=3\).

\(\displaystyle\int_{1}^{4}\dfrac{6}{2x+1}\,dx\)	\(=\)	\(3\Big[\ln(2x+1)\Big]_{1}^{4}\)
	\(=\)	\(3(\ln 9-\ln 3)=3\ln 3\)

3 \ln 3

Example 4 — Area under a curve

Find the area under \(y=2e^{x}\) from \(x=0\) to \(x=1\).

Solution

The region is above the axis, so the area is the definite integral.

\(\text{Area}\)	\(=\)	\(\displaystyle\int_{0}^{1} 2e^{x}\,dx=\Big[2e^{x}\Big]_{0}^{1}\)
	\(=\)	\(2e-2=2(e-1)\approx 3.44\)

2 (e - 1)

Common pitfalls

Divide by the coefficient of \(x\). \(\displaystyle\int e^{2x}\,dx=\dfrac{1}{2}e^{2x}+C\), not \(e^{2x}\) or \(2e^{2x}\); the same \(\tfrac{1}{a}\) factor applies to \((ax+b)^{r}\) and \(\dfrac{1}{ax+b}\).

The reciprocal gives a logarithm. \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\), because the power rule fails at \(n=-1\). Do not write \(\dfrac{x^{0}}{0}\).

Mind the sign and never drop \(+C\). \(\displaystyle\int \sin x\,dx=-\cos x+C\) (minus), \(\displaystyle\int \cos x\,dx=\sin x+C\) (no minus). Keep \(+C\) until a boundary condition fixes it or you evaluate a definite integral.

Frequently asked questions

What are the standard integrals in Methods?

\(\displaystyle\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}+C\ (n\neq-1)\), \(\displaystyle\int e^{x}\,dx=e^{x}+C\), \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\), \(\displaystyle\int \sin x\,dx=-\cos x+C\) and \(\displaystyle\int \cos x\,dx=\sin x+C\).

How do you integrate e^{kx} or (ax+b)^r?

Integrate as if the inner were \(x\), then divide by the coefficient \(a\). So \(\displaystyle\int e^{kx}\,dx=\dfrac{1}{k}e^{kx}+C\) and \(\displaystyle\int (ax+b)^{r}\,dx=\dfrac{(ax+b)^{r+1}}{a(r+1)}+C\).

Why does 1/x integrate to a logarithm?

The power rule divides by \(n+1\), which is \(0\) at \(n=-1\), so it fails. Instead \(\displaystyle\int \dfrac{1}{x}\,dx=\ln|x|+C\), since \(\dfrac{d}{dx}\ln|x|=\dfrac{1}{x}\).

How do you integrate a sum of several forms?

By linearity: integrate each term with its own standard form and keep one \(+C\). E.g. \(\displaystyle\int (6x^{2}+e^{2x}-\cos 3x)\,dx=2x^{3}+\dfrac{1}{2}e^{2x}-\dfrac{1}{3}\sin 3x+C\).

How do you evaluate a definite integral?

Anti-differentiate (no \(+C\)), then compute \(F(b)-F(a)\) with exact values. E.g. \(\displaystyle\int_{1}^{e}\dfrac{1}{x}\,dx=\big[\ln x\big]_{1}^{e}=1-0=1\).

How is a definite integral related to area?

If \(f(x)\ge 0\) on \([a,b]\), the area between the curve and the \(x\)-axis is \(\displaystyle\int_{a}^{b} f(x)\,dx\). The same idea gives a total change from a rate.

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