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:
- $F\left(x\right)=3x^2+4x+5$F(x)=3x2+4x+5
- $F\left(x\right)=3x^2+4x+1$F(x)=3x2+4x+1
- $F\left(x\right)=3x^2+4x-17$F(x)=3x2+4x−17
- $F\left(x\right)=3x^2+4x-3$F(x)=3x2+4x−3
- $F\left(x\right)=3x^2+4x+a$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)=naxn−1. 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.
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=1x3−√x, 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=x−3−x12
| $y$y | $=$= | $\int x^{-3}-x^{\frac{1}{2}}\ dx$∫x−3−x12 dx |
| $=$= | $\frac{x^{-2}}{-2}-x^{\frac{3}{2}}\div\frac{3}{2}+c$x−2−2−x32÷32+c | |
| $=$= | $-\frac{1}{2x^2}-\frac{2x^{\frac{3}{2}}}{3}+c$−12x2−2x323+c, where $c$c is a constant. |
Example 3
Given that $f'(x)=6x^2-4x+1$f′(x)=6x2−4x+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) | $=$= | $\int6x^2-4x+1\ dx$∫6x2−4x+1 dx |
| $=$= | $\frac{6x^3}{3}-\frac{4x^2}{2}+x+c$6x33−4x22+x+c | |
| $=$= | $2x^3-2x^2+x+c$2x3−2x2+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)3−2(2)2+(2)+c | $=$= | $7$7 |
| $16-8+2+c$16−8+2+c | $=$= | $7$7 |
| $\therefore c$∴c | $=$= | $-3$−3 |
Hence, $f(x)=2x^3-2x^2+x-3$f(x)=2x3−2x2+x−3.
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 $\int\left(\frac{x^3}{3}-\frac{3}{x^3}\right)dx$∫(x33−3x3)dx, with $C$C as the constant of integration.
Question 4
Suppose $\frac{dy}{dx}=9x^2-10x+2$dydx=9x2−10x+2.
Find an equation for $y$y.
Use $C$C as the constant of integration.
Solve for $C$C if the curve passes through the point $\left(2,13\right)$(2,13).
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)=5t2−2t+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) | $=$= | $\int5t^2-2t+3\ dt$∫5t2−2t+3 dt |
| $=$= | $\frac{5t^3}{3}-t^2+3t+c$5t33−t2+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)=5t33−t2+3t−2.
(b) Substitute $t=3$t=3 into the displacement function, we obtain:
| $s(3)$s(3) | $=$= | $\frac{5\times27}{3}-9+9-2$5×273−9+9−2 |
| $=$= | $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.
Calculate the initial velocity of the particle.
Determine the function $x\left(t\right)$x(t) for the position of the particle. Use $C$C as the constant of integration.
Calculate the displacement of the particle after four seconds.
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)=12t2−48t, where $t\ge0$t≥0.
The object starts its movement $5$5 feet to the right of the origin. That is, $x\left(0\right)=5$x(0)=5.
State the displacement $x\left(t\right)$x(t) of the particle at time $t$t. Use $C$C as the constant of integration.
Solve for the times $t$t when the object is at rest.
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)=6t−27, where $t\ge0$t≥0.
After $10$10 seconds, the object is moving at $90$90 metres per second in the positive direction.
State the velocity $v$v of the particle at time $t$t. Use $C$C as the constant of integration.
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.