The antiderivative and the indefinite integral

Integral calculus, grade 11 (5 units) Lesson 1 of 20Learning

The antiderivative and the indefinite integral

An antiderivative

In grade 10 you learnt: an antiderivative of $\displaystyle f$ is a function $\displaystyle F$ whose derivative is $\displaystyle f$ , that is $\displaystyle {F'(x)}={f(x)}$ .

$\displaystyle {F(x)}={x^3}$ is an antiderivative of $\displaystyle {f(x)}={3x^2}$ , because $\displaystyle {(x^3)'}={3x^2}$ .
$\displaystyle x^3+5$ and $\displaystyle x^3-2$ are antiderivatives of $\displaystyle 3x^2$ too: the derivative of a constant is 0.
All the antiderivatives of $\displaystyle f$ have the same slope at every $\displaystyle x$ , so their graphs are one graph, shifted up or down.
The antiderivatives of $\displaystyle f$ are all $\displaystyle F(x)+C$ , where $\displaystyle C$ is any number.

\displaystyle y=x^{3}

\displaystyle y=x^{3}+2

\displaystyle y=x^{3}-2

The indefinite integral

All the antiderivatives of $\displaystyle f$ are written with one sign, the indefinite integral of $\displaystyle f$ :

\displaystyle \int f(x)\,dx=F(x)+C

\displaystyle F'(x)=f(x)

Read: “the integral of $\displaystyle f(x)$ with respect to $\displaystyle x$ ”.
$\displaystyle dx$ says that $\displaystyle x$ is the variable.
$\displaystyle C$ is the constant of integration: without it only one antiderivative of the family is left.
Finding the integral is called integration, and it is the reverse of differentiating. For example, $\displaystyle {\int3x^2\,dx}={x^3+C}$ .

The integral of a power

Differentiating $\displaystyle x^{n+1}$ gives $\displaystyle (n+1)x^n$ . So to get $\displaystyle x^n$ , raise the power by 1 and divide by the new power:

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

\displaystyle n\neq-1

\displaystyle \int k\,dx=kx+C

$\displaystyle {\int x^4\,dx}={\frac{x^5}{5}+C}$ .
$\displaystyle {\int x\,dx}={\frac{x^2}{2}+C}$ .
$\displaystyle {\int7\,dx}={7x+C}$ .
The check: differentiate the result, and you must get the function inside the integral.

The integral of a polynomial

Integrate a polynomial term by term: a number multiplying a power keeps multiplying it, and the integrals of the terms are added. The example below finds $\displaystyle \int\left(6x^2-4x+5\right)dx$ :

$\displaystyle {\int6x^2\,dx}={6\cdot\frac{x^3}{3}}={2x^3}$ .
$\displaystyle {\int4x\,dx}={4\cdot\frac{x^2}{2}}={2x^2}$ .
$\displaystyle {\int5\,dx}={5x}$ .
One constant, $\displaystyle C$ , for the whole integral:

\int\left(6x^2-4x+5\right)dx=2x^3-2x^2+5x+C

The check: $\displaystyle {\left(2x^3-2x^2+5x+C\right)'}={6x^2-4x+5}$ .

A product: expand first

There is no rule for the integral of a product: the integral of a product is not the product of the integrals. So expand the brackets first. The example below finds $\displaystyle \int x(3x-2)\,dx$ :

\int x(3x-2)\,dx=\int\left(3x^2-2x\right)dx=x^3-x^2+C

The product of the integrals, $\displaystyle \frac{x^2}{2}\cdot\left(\frac{3x^2}{2}-2x\right)$ , differentiates to a polynomial of degree three – not to $\displaystyle 3x^2-2x$ .

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

The variable here is $\displaystyle t$ .

/ – 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

$\displaystyle {F(x)}={x^3+2x^2}$ is an antiderivative of a function $\displaystyle f$ .
Find $\displaystyle f(x)$ .

/ – 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 $\displaystyle {f(x)}={3x^2-3}$ is given.

Find all the antiderivatives of $\displaystyle f$ .

/ – 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

The graph of one of the antiderivatives passes through the point $\displaystyle A(0,2)$ .
Write it on the board.

\displaystyle A=(0,2)

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