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10.06 Anti-differentiation

Lesson

We can use differentiation to find the derivative, which we have seen has many useful applications. What if we were given a derivative function, can we go backwards to find the original function it came from? Can we "anti-differentiate"? 

Exploration

If we know that the derivative of some function $f(x)$f(x), is $6x+4$6x+4 can we determine what $f(x)$f(x) is? What degree is $f(x)$f(x)? Do we have enough information?

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?

As all the constant terms $5,1,-17,-3$5,1,17,3 and $a$a have a derivative of zero, we have the same gradient function 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 and we are looking for the original function, how do we know which one it is? Without extra information, such as a point the original function passes through, we cannot tell what the constant term may have been.   

In fact, when we go 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 that we don't yet know. When including the constant we often refer to the form as the primitive function, general anti-derivative or 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, our differentiation rule tells us $f'(x)=nax^{n-1}$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 we want to "undo" that.  Reversing the rule we get "add $1$1 to the power of $x$x, then divide by 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 known 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

Determine the anti-derivative $F\left(x\right)$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

Find $\int\left(\frac{x^3}{3}-\frac{3}{x^3}\right)dx$(x333x3)dx, with $C$C as the constant of integration.

Question 4

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 we have a function that 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, we obtain displacement as a function of time 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 are 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 that 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 5

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 6

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 7

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

2.3.22

calculate anti-derivatives of polynomial functions

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