Antidifferentiation

We have already seen how to find the antiderivative of some very simple polynomials

For example, the antiderivative of a constant term $a$a, is $ax+C$ax+C

We can also see some examples of powers of $x$x, 

• Antiderivative of $x$x is $\frac{x^2}{2}+C$x22​+C
• Antiderivative of $x^2$x2 is $\frac{x^3}{3}+C$x33​+C
• Antiderivative of $x^3$x3 is $\frac{x^4}{4}+C$x44​+C

Which leads to a general form of 

• Antiderivative of $x^n$xn is $\frac{x^{n+1}}{n+1}+C$xn+1n+1​+C

 

Formal notation

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$∫xndx=xn+1n+1​+C

 

Properties

• The antiderivative of a sum is the sum of the antiderivatives.

$\int af(x)\pm bg(x)dx=a\int f(x)dx\pm b\int g(x)dx$∫af(x)±bg(x)dx=a∫f(x)dx±b∫g(x)dx

• Integral of a constant  $\int adx=ax+C$∫adx=ax+C
• All functions that are continuous at all points in an interval have antiderivatives that are defined on the interval. 

 

Using the properties

To find indefinite integrals of polynomial functions we combine the idea that the antiderivative of a sum is the same as the sum of the anitderivatives and our rule for finding the antiderivative of a power term together.

 

Example

Find $$

$\int3x^4+4x^3-2x^2+x-7dx$∫3x4+4x3−2x2+x−7dx	$=$=	$\int3x^4dx+\int4x^3dx-\int2x^2dx+\int xdx-\int7dx$∫3x4dx+∫4x3dx−∫2x2dx+∫xdx−∫7dx
	$=$=	$3\int x^4dx+4\int x^3dx-2\int x^2dx+\int xdx-\int7dx$3∫x4dx+4∫x3dx−2∫x2dx+∫xdx−∫7dx
	$=$=	$3\times\frac{x^5}{5}+4\times\frac{x^4}{4}-2\times\frac{x^3}{3}+\frac{x^2}{2}-7x+C$3×x55​+4×x44​−2×x33​+x22​−7x+C
	$=$=	$\frac{3x^5}{5}+\frac{4x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}-7x+C$3x55​+4x44​−2x33​+x22​−7x+C
	$=$=	$\frac{3x^5}{5}+x^4-\frac{2x^3}{3}+\frac{x^2}{2}-7x$3x55​+x4−2x33​+x22​−7x

 

Example

A particular curve has a derivative given by $\frac{dy}{dx}=2x^4-3x^2+x$dydx​=2x4−3x2+x

We are told that the curve passes through the point $(1,-\frac{1}{2})$(1,−12​).  Find the equation of the curve. 

We start here with the indefinite integral

$\int2x^4-3x^2+xdx$∫2x4−3x2+xdx	$=$=	$2\int x^4dx-3\int x^2dx+\int xdx$2∫x4dx−3∫x2dx+∫xdx
	$=$=	$2\cdot\frac{x^5}{5}-3\cdot\frac{x^3}{3}+\frac{x^2}{2}+C$2·x55​−3·x33​+x22​+C
	$=$=	$\frac{2x^5}{5}-x^3+\frac{x^2}{2}+C$2x55​−x3+x22​+C

 

We are given a set of conditions that can help us identify the value of $C$C.  We use the point $(1,-\frac{1}{2})$(1,−12​)

$y$y	$=$=	$\frac{2x^5}{5}-x^3+\frac{x^2}{2}+C$2x55​−x3+x22​+C
$-\frac{1}{2}$−12​	$=$=	$\frac{2}{5}-1+\frac{1}{2}+C$25​−1+12​+C
$-\frac{1}{2}-\frac{2}{5}+1-\frac{1}{2}$−12​−25​+1−12​	$=$=	$C$C
$C$C	$=$=	$-\frac{2}{5}$−25​

 

So the curve is $y=\frac{2x^5}{5}-x^3+\frac{x^2}{2}-\frac{2}{5}$y=2x55​−x3+x22​−25​

 

Worked Examples

Question 1

Question 2

Question 3