The integral of a composite function with a linear inner function

Integral calculus, grade 11 (5 units) Lesson 10 of 20Learning

The integral of a composite function with a linear inner function

The rule

In grade 10 you learnt the chain rule: $\displaystyle {\left(F(mx+b)\right)'}={F'(mx+b)\cdot m}={m\cdot f(mx+b)}$ . The inner derivative $\displaystyle m$ is a number, so to get $\displaystyle f(mx+b)$ without the $\displaystyle m$ , divide by $\displaystyle m$ :

\displaystyle \int f(mx+b)\,dx=\frac{1}{m}F(mx+b)+C

\displaystyle m\neq0

$\displaystyle F$ is an antiderivative of $\displaystyle f$ .
That is: find the integral as if $\displaystyle mx+b$ were $\displaystyle x$ , and divide by the coefficient of $\displaystyle x$ .

A power, a root and a fraction

In the examples below the inner function is linear, and the rule gives the integral:

$\displaystyle \int(3x-1)^4\,dx$ : the integral of $\displaystyle u^4$ is $\displaystyle \frac{u^5}{5}$ , and $\displaystyle {m}={3}$ :

\int(3x-1)^4\,dx=\frac{1}{3}\cdot\frac{(3x-1)^5}{5}+C=\frac{(3x-1)^5}{15}+C

$\displaystyle \int\sqrt{2x+1}\,dx$ : the integral of $\displaystyle \sqrt{u}$ is $\displaystyle \frac{2}{3}u\sqrt{u}$ , and $\displaystyle {m}={2}$ :

\int\sqrt{2x+1}\,dx=\frac{1}{2}\cdot\frac{2}{3}(2x+1)\sqrt{2x+1}+C=\frac{(2x+1)\sqrt{2x+1}}{3}+C

$\displaystyle \int\frac{1}{(5-x)^2}\,dx$ : the integral of $\displaystyle \frac{1}{u^2}$ is $\displaystyle -\frac{1}{u}$ , and $\displaystyle {m}={-1}$ :

\int\frac{1}{(5-x)^2}\,dx=\frac{1}{-1}\cdot\left(-\frac{1}{5-x}\right)+C=\frac{1}{5-x}+C

The check, for the first one say: $\displaystyle {\left(\frac{(3x-1)^5}{15}\right)'}={\frac{5(3x-1)^4\cdot3}{15}}={(3x-1)^4}$ .

A definite integral

The example below calculates $\displaystyle \int_0^1(2x+1)^3\,dx$ by the same rule, and puts the limits into the antiderivative:

\int_0^1(2x+1)^3\,dx=\left[\frac{(2x+1)^4}{8}\right]_0^1=\frac{81}{8}-\frac{1}{8}=10

An inner function that is not linear

The rule holds only when the inner function is linear. For $\displaystyle \int(x^2+1)^2\,dx$ the “rule” gives $\displaystyle \frac{(x^2+1)^3}{3\cdot2x}$ , but the derivative of that expression is not $\displaystyle (x^2+1)^2$ : $\displaystyle 2x$ is not a number, and it cannot be taken out of the differentiation. Here, expand the brackets:

\int(x^2+1)^2\,dx=\int\left(x^4+2x^2+1\right)dx=\frac{x^5}{5}+\frac{2x^3}{3}+x+C

Exercise 1

Find the integrals.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

Calculate the integrals.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

The function $\displaystyle {f(x)}={\frac{1}{\sqrt{x+4}}}$ is given.
$\displaystyle {F(x)}={\int_0^x f(t)\,dt}$ .

Write $\displaystyle F(x)$ without an integral.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Find the $\displaystyle x$ for which $\displaystyle {F(x)}={2}$ .

Formulas: integral calculus, grade 11 (5 units)

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Basic integrals

The indefinite integral – all the antiderivatives
$\displaystyle \int f(x)\,dx=F(x)+C$ $\displaystyle F^{\prime}(x)=f(x)$
A power
$\displaystyle \int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$ $\displaystyle n\neq-1$
A constant
$\displaystyle \int k\,dx=kx+C$
$\displaystyle \int\frac{1}{x^{2}}\,dx=-\frac{1}{x}+C$
One over a root
$\displaystyle \int\frac{1}{\sqrt{x}}\,dx=2\sqrt{x}+C$
A root
$\displaystyle \int\sqrt{x}\,dx=\frac{2}{3}x\sqrt{x}+C$

The rules of integration

A constant factor
$\displaystyle \int k\cdot f(x)\,dx$ $\displaystyle =$ $\displaystyle k\int f(x)\,dx$
A sum and a difference
$\displaystyle \int\left(f(x)\pm g(x)\right)dx$ $\displaystyle =$ $\displaystyle \int f(x)\,dx\pm\int g(x)\,dx$
A linear inner function
$\displaystyle \int f(mx+b)\,dx=\frac{1}{m}F(mx+b)+C$ $\displaystyle m\neq0$
A composite function times the inner derivative
$\displaystyle \int f(u(x))\cdot u^{\prime}(x)\,dx=F(u(x))+C$
A power of a function times its derivative
$\displaystyle \int\left(u(x)\right)^{n}\cdot u^{\prime}(x)\,dx=\frac{\left(u(x)\right)^{n+1}}{n+1}+C$ $\displaystyle n\neq-1$

Trigonometric functions (in radians)

$\displaystyle \int\cos x\,dx=\sin x+C$
$\displaystyle \int\sin x\,dx=-\cos x+C$
$\displaystyle \int\frac{1}{\cos^{2}x}\,dx=\tan x+C$
From $\displaystyle {\cos2x}={1-2\sin^2x}$
$\displaystyle \sin^{2}x$ $\displaystyle =$ $\displaystyle \frac{1-\cos2x}{2}$
From $\displaystyle {\cos2x}={2\cos^2x-1}$
$\displaystyle \cos^{2}x$ $\displaystyle =$ $\displaystyle \frac{1+\cos2x}{2}$
From $\displaystyle {\sin2x}={2\sin x\cos x}$
$\displaystyle \sin x\cos x$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sin2x$
From $\displaystyle {\sin^2x+\cos^2x}={1}$
$\displaystyle \tan^{2}x$ $\displaystyle =$ $\displaystyle \frac{1}{\cos^{2}x}-1$

The accumulation function and the definite integral

The accumulation function of $\displaystyle f$ from $\displaystyle a$
$\displaystyle F_{a}(x)=\int_{a}^{x}f(t)\,dt$
The fundamental theorem
$\displaystyle F_{a}^{\prime}(x)=f(x)$
The definite integral, $\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int_{a}^{b}f(x)\,dx=F(b)-F(a)$
The same limit
$\displaystyle \int_{a}^{a}f(x)\,dx=0$
Swapping the limits
$\displaystyle \int_{b}^{a}f(x)\,dx$ $\displaystyle =$ $\displaystyle -\int_{a}^{b}f(x)\,dx$
Joining intervals
$\displaystyle \int_{a}^{b}f(x)\,dx+\int_{b}^{c}f(x)\,dx$ $\displaystyle =$ $\displaystyle \int_{a}^{c}f(x)\,dx$

Areas and accumulation

An area above the $\displaystyle x$ -axis
$\displaystyle S=\int_{a}^{b}f(x)\,dx$ $\displaystyle f(x)\geq0$
An area below the $\displaystyle x$ -axis
$\displaystyle S=-\int_{a}^{b}f(x)\,dx$ $\displaystyle f(x)\leq0$
The area between two graphs
$\displaystyle S=\int_{a}^{b}\left(f(x)-g(x)\right)dx$ $\displaystyle f(x)\geq g(x)$
The displacement from the velocity
$\displaystyle s(b)-s(a)=\int_{a}^{b}v(t)\,dt$