Integrals of negative powers and of roots

Integral calculus, grade 11 (5 units) Lesson 2 of 20Learning

Integrals of negative powers and of roots

A negative power

The rule $\displaystyle {\int x^n\,dx}={\frac{x^{n+1}}{n+1}+C}$ holds for every $\displaystyle n\neq-1$ , negative too. To use it, write a fraction with $\displaystyle x$ in the denominator as a negative power: $\displaystyle {\frac{1}{x^n}}={x^{-n}}$ . The example below finds $\displaystyle \int\frac{1}{x^2}\,dx$ :

\int\frac{1}{x^2}\,dx=\int x^{-2}\,dx=\frac{x^{-1}}{-1}+C=-\frac{1}{x}+C

The check: $\displaystyle {\left(-\frac{1}{x}\right)'}={\frac{1}{x^2}}$ .
The function is defined only where $\displaystyle x\neq0$ .
$\displaystyle {n}={-1}$ , that is $\displaystyle \frac{1}{x}$ , is the exception: the denominator $\displaystyle n+1$ would be 0. You will learn its integral in grade 12.

\displaystyle \int\frac{1}{x^{2}}\,dx=-\frac{1}{x}+C

A root

In grade 10 you learnt that $\displaystyle {\sqrt{x}}={x^{\frac{1}{2}}}$ . The same power rule gives the integrals of roots:

\int\sqrt{x}\,dx=\int x^{\frac{1}{2}}\,dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{2}{3}x\sqrt{x}+C

\int\frac{1}{\sqrt{x}}\,dx=\int x^{-\frac{1}{2}}\,dx=\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C=2\sqrt{x}+C

$\displaystyle {x^{\frac{3}{2}}}={x\cdot x^{\frac{1}{2}}}={x\sqrt{x}}$ .
The check: $\displaystyle {\left(2\sqrt{x}\right)'}={2\cdot\frac{1}{2\sqrt{x}}}={\frac{1}{\sqrt{x}}}$ .

\displaystyle \int\sqrt{x}\,dx=\frac{2}{3}x\sqrt{x}+C

\displaystyle \int\frac{1}{\sqrt{x}}\,dx=2\sqrt{x}+C

A fraction: split it into a sum

When the denominator of a fraction is a single term, divide each term of the numerator by it, and you get a sum of powers. The example below finds $\displaystyle \int\frac{x^3-2}{x^2}\,dx$ :

$\displaystyle {\frac{x^3-2}{x^2}}={\frac{x^3}{x^2}-\frac{2}{x^2}}={x-2x^{-2}}$ .
$\displaystyle {\int\left(x-2x^{-2}\right)dx}={\frac{x^2}{2}-2\cdot\frac{x^{-1}}{-1}+C}={\frac{x^2}{2}+\frac{2}{x}+C}$ .

In the same way, $\displaystyle {\frac{x+1}{\sqrt{x}}}={\frac{x}{\sqrt{x}}+\frac{1}{\sqrt{x}}}={\sqrt{x}+\frac{1}{\sqrt{x}}}$ , so:

\int\frac{x+1}{\sqrt{x}}\,dx=\frac{2}{3}x\sqrt{x}+2\sqrt{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

$\displaystyle {\sqrt[3]{x}}={x^{\frac{1}{3}}}$ .

/ – 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

Split each fraction into a sum, and find the 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

Exercise 3

The derivative of a function $\displaystyle f$ is $\displaystyle {f'(x)}={\frac{1}{\sqrt{x}}}$ , and $\displaystyle x>0$ .
The graph of $\displaystyle f$ passes through the point $\displaystyle A(4,3)$ .

Find $\displaystyle C$ .

Write $\displaystyle f$ on the board.

\displaystyle A=(4,3)

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
$\displaystyle s(b)-s(a)=\int_{a}^{b}v(t)\,dt$