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VCE 11 Methods 2023

10.06 Anti-differentiation

Lesson

Differentiation can be used to find the derivative, which has many useful applications. If given a derivative function, is it possible to go backwards to find the original function it came from? Is it possible to "anti-differentiate"? 

For example, if the derivative of some function $f(x)$f(x) was $6x+4$6x+4, what is $f(x)$f(x)? What degree is $f(x)$f(x)? Is there enough information to solve this problem?

Let's explore this problem by first finding the derivative of the following $5$5 functions:

  1. $F\left(x\right)=3x^2+4x+5$F(x)=3x2+4x+5
  2. $F\left(x\right)=3x^2+4x+1$F(x)=3x2+4x+1
  3. $F\left(x\right)=3x^2+4x-17$F(x)=3x2+4x17
  4. $F\left(x\right)=3x^2+4x-3$F(x)=3x2+4x3
  5. $F\left(x\right)=3x^2+4x+a$F(x)=3x2+4x+a

What did you notice?

Well, because all the constant terms $5,1,-17,-3$5,1,17,3 and $a$a all had a derivative of zero, the same gradient function was found for all of them: $F'\left(x\right)=6x+4$F(x)=6x+4

So, given the gradient function of $f(x)$f(x) is $6x+4$6x+4, if looking for the original function, how would we know which one it was? Without extra information, such as a point the original function passes through, it's impossible to tell what the constant term may have been.   

In fact, when "going backwards" using anti-differentiation, we cannot be sure which function was the original so we find what is called a primitive function. We can give the form of all possible primitive functions by adding an unknown constant.

For example, all primitive functions of $6x+4$6x+4 have the form $F\left(x\right)=3x^2+4x+C$F(x)=3x2+4x+C, where $C$C is a constant term which is unknown. When including the constant, this form is referred to as the primitive function and is also known as the general anti-derivative or the indefinite integral.

The process of "going backwards" from differentiation to find a primitive function is called anti-differentiation or indefinite integration.

For power functions of the form $f(x)=ax^n$f(x)=axn, by differentiation $f'(x)=nax^{n-1}$f(x)=naxn1. That is, "multiply the expression by the power of $x$x and then take $1$1 from the power." For anti-differentiation, this rule is "undone".  Reversing the rule gives "add $1$1 to the power of $x$x, then divide the expression by the new power".

So, for $f(x)=ax^n$f(x)=axn, the indefinite integral is $F(x)=\frac{ax^{n+1}}{n+1}+C$F(x)=axn+1n+1+C. The value $C$C is know as the constant of integration.

Notation: The expression $\int f(x)\ dx$f(x) dx, reads "anti-differentiate the function $f(x)$f(x) with respect to $x$x.  For example, $\int3x^2dx=x^3+C$3x2dx=x3+C, where $C$C is a constant, reads as the general anti-derivative of $3x^2$3x2 with respect to $x$x is $x^3+C$x3+C.

Summary
For a function $f(x)$f(x), if $F'(x)=f(x)$F(x)=f(x) then $F(x)$F(x) in an anti-derivative or primitive function of $f(x)$f(x)
The indefinite integral of $f(x)$f(x) is given by:
$\int f(x)\ dx=F(x)+C$f(x) dx=F(x)+C, where $C$C is a constant called the constant of integration.
For functions of the form $f(x)=ax^n$f(x)=axn:
$\int ax^n\ dx=\frac{ax^{n+1}}{n+1}+C$axn dx=axn+1n+1+C, where $C$C is a constant
Linearity properties of the integral means we can integrate a function term be term. That is:
$\int\left[f(x)+g(x)\right]\ dx=\int f(x)\ dx+\int g(x)\ dx$[f(x)+g(x)] dx=f(x) dx+g(x) dx

 

Worked examples

Example 1

Find the primitive function $F\left(x\right)$F(x) when $f\left(x\right)=3x^2+4x$f(x)=3x2+4x.

Think: Term by term, we raise the power by one and divide by the new power.

Do:

$F(x)$F(x) $=$= $\frac{3x^{2+1}}{2+1}+\frac{4x^2}{2}+C$3x2+12+1+4x22+C Don't forget the constant of integration
  $=$= $\frac{3x^3}{3}+2x^2+C$3x33+2x2+C  
  $=$= $x^3+2x^2+C$x3+2x2+C, where $C$C is a constant.  
Example 2

Given that $\frac{dy}{dx}=\frac{1}{x^3}-\sqrt{x}$dydx=1x3x, find the form of all anti-derivatives $y$y.

Think: First rewrite all terms as powers of $x$x and then integrate term by term-.   

Do: $\frac{dy}{dx}=x^{-3}-x^{\frac{1}{2}}$dydx=x3x12

$y$y $=$= $\int x^{-3}-x^{\frac{1}{2}}\ dx$x3x12 dx
  $=$= $\frac{x^{-2}}{-2}-x^{\frac{3}{2}}\div\frac{3}{2}+C$x22x32÷32+C
  $=$= $-\frac{1}{2x^2}-\frac{2x^{\frac{3}{2}}}{3}+C$12x22x323+C, where $C$C is a constant.
Example 3

Given that $f'(x)=6x^2-4x+1$f(x)=6x24x+1 and $f(x)$f(x) passes through the point $\left(2,7\right)$(2,7), find the function $f(x)$f(x).

Think: Find the indefinite integral and then use the given point to find the constant for the given function. 

Do: 

$f(x)$f(x) $=$= $\int6x^2-4x+1\ dx$6x24x+1 dx
  $=$= $\frac{6x^3}{3}-\frac{4x^2}{2}+x+C$6x334x22+x+C
  $=$= $2x^3-2x^2+x+C$2x32x2+x+C

Given $f(x)$f(x) passes through $\left(2,7\right)$(2,7), we have:

$f(2)$f(2) $=$= $7$7
Thus, $2\left(2\right)^3-2\left(2\right)^2+\left(2\right)+C$2(2)32(2)2+(2)+C $=$= $7$7
$16-8+2+C$168+2+C $=$= $7$7
$\therefore C$C $=$= $-3$3

Hence, $f(x)=2x^3-2x^2+x-3$f(x)=2x32x2+x3.

 

Practice questions

Question 1

Find the primitive function of $15x^4+16x^3$15x4+16x3.

Use $C$C as the constant of integration.

Question 2

Given that $\frac{dy}{dx}=\sqrt{x}$dydx=x, find the primitive function $y$y.

Use $C$C as the constant of integration.

Question 3

Suppose $\frac{dy}{dx}=9x^2-10x+2$dydx=9x210x+2.

  1. Find an equation for $y$y.

    Use $C$C as the constant of integration.

  2. Solve for $C$C if the curve passes through the point $\left(2,13\right)$(2,13).

  3. Hence find the equation of $y$y.

 

Applications of anti-differentiation

Generally, if a function gives the rate of change of some quantity, often with respect to time, then an anti-derivative will be a function of the quantity itself. 

For example, velocity is the rate of change of displacement with respect to time. So, given velocity as a function of time, displacement as a function of time will be obtained by anti-differentiation.

Similarly, acceleration is the rate of change of velocity with respect to time. So, a velocity function is obtained from an acceleration function by anti-differentiation..

 Thus for displacement $x$x (sometimes $s$s), velocity $v$v and acceleration $a$a, we have: 

$x(t)=\int v(t)\ dt$x(t)=v(t) dt

$v(t)=\int a(t)\ dt$v(t)=a(t) dt

 

Worked examples

Example 4

Consider a body moving so that its velocity, in $m/s$m/s, as a function of time is given by $v(t)=5t^2-2t+3$v(t)=5t22t+3. Given that its displacement at $t=0$t=0 is $-2$2 $m$m:

a) Find its displacement function $s(t)$s(t).

b) What is the displacement at $t=3$t=3?

a) Since $s(t)=\int v(t)\ dt$s(t)=v(t) dt, we have:

$s(t)$s(t) $=$= $\int5t^2-2t+3\ dt$5t22t+3 dt
  $=$= $\frac{5t^3}{3}-t^2+3t+C$5t33t2+3t+C

We know that $s(0)=-2$s(0)=2. Therefore, $C=-2$C=2 and the displacement function is given by $s(t)=\frac{5t^3}{3}-t^2+3t-2$s(t)=5t33t2+3t2.

b) Substitute $t=3$t=3 into the displacement function, we obtain:

$s(3)$s(3) $=$= $\frac{5\times27}{3}-9+9-2$5×2739+92
  $=$= $43$43

The object is $43$43 $m$m to the right of the origin.

Example 5

A falling body near the surface of the earth has a constant acceleration of about $-9.8m/s^2$9.8m/s2. If the body starts from rest at $t=0$t=0, find its velocity function. Given that the body begins its fall from a height of $200m$200m, find its displacement function.

Acceleration is the rate of change of velocity. So using $v(t)=\int a(t)\ dt$v(t)=a(t) dt, we have:

$v(t)$v(t) $=$= $\int(-9.8)\ dt$(9.8) dt
  $=$= $-9.8t+c$9.8t+c

 

We were given that the body started from rest, that is it had an initial velocity of $0m/s$0m/s. So we have:

$v(0)$v(0) $=$= $-9.8\times0+c$9.8×0+c
$0$0 $=$= $c$c

Hence, the velocity function is $v(t)=-9.8t$v(t)=9.8t.

Using the fact that the displacement is given by $x(t)=\int v(t)\ dt$x(t)=v(t) dt, we obtain:

$x(t)$x(t) $=$= $\int-9.8t\ dt$9.8t dt
  $=$= $\frac{-9.8t^2}{2}+d$9.8t22+d
  $=$= $-4.9t^2+d$4.9t2+d

Using the information at $t=0$t=0, $s=200$s=200. We can see $d=200$d=200 and the displacement function is $s(t)=-4.9t^2+200$s(t)=4.9t2+200.

 

Practice questions

Question 4

The velocity of a particle moving in a straight line is given by $v\left(t\right)=6t+15$v(t)=6t+15, where $v$v is the velocity in metres per second and $t$t is the time in seconds.

The displacement after $2$2 seconds is $45$45 metres to the right of the origin.

  1. Calculate the initial velocity of the particle.

  2. Determine the function $x\left(t\right)$x(t) for the position of the particle. Use $C$C as the constant of integration.

  3. Calculate the displacement of the particle after four seconds.

  4. Find the total distance travelled between two and four seconds.

Question 5

The velocity $v\left(t\right)$v(t) of an object travelling horizontally along a straight line after $t$t seconds is modelled by $v\left(t\right)=12t^2-48t$v(t)=12t248t, where $t\ge0$t0.

The object starts its movement $5$5 feet to the right of the origin. That is, $x\left(0\right)=5$x(0)=5.

  1. State the displacement $x\left(t\right)$x(t) of the particle at time $t$t. Use $C$C as the constant of integration.

  2. Solve for the times $t$t when the object is at rest.

  3. Find the displacement at which the object is stationary other than its initial position.

Question 6

The acceleration $a$a (in m/s2) of an object travelling horizontally along a straight line after $t$t seconds is modelled by $a\left(t\right)=6t-27$a(t)=6t27, where $t\ge0$t0.

After $10$10 seconds, the object is moving at $90$90 metres per second in the positive direction.

  1. State the velocity $v$v of the particle at time $t$t. Use $C$C as the constant of integration.

  2. Solve for all times $t$t at which the particle is at rest. If there are multiple solutions write them on the same line separated by commas.

 

Outcomes

U2.AoS3.13

find by hand the derivative function and an anti-derivative function for a polynomial function of low degree

U2.AoS3.6

anti-differentiation as the inverse process of differentiation and identification of families of curves with the same gradient function, and the use of a boundary condition to determine a specific anti-derivative of a given function.

U2.AoS3.15

find a family of anti-derivative functions for a given polynomial function, and determine a specific antiderivative given a boundary condition

U2.AoS3.10

the rules for finding derivatives and anti-derivatives of simple polynomial functions

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