iGCSE (2021 Edition)

# 12.03 Finding f(x) from f'(x)

Lesson

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

We have seen that anti-derivatives are also known as primitive functions. The process of anti-differentiation gives a family of primitive functions $f(x)+C$f(x)+C that share the same derivative. This occurs because the derivative of a constant is zero. We can determine the value of the constant $C$C if we are given more information such as the coordinates of a point on the function.

#### Worked example

##### Example 1

A particular curve has a derivative given by $\frac{dy}{dx}=2x^4-3x^2+x$dydx=2x43x2+x. The curve passes through the point $(1,-\frac{1}{2})$(1,12).  Find the equation of the curve.

Think: Our starting point is to find a primitive function by integrating, using $C$C as the constant.

Do: We start here with the indefinite integral:

 $\int2x^4-3x^2+xdx$∫2x4−3x2+xdx $=$= $2\times\frac{x^5}{5}-3\times\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​

Hence the equation of the curve is:

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

#### Practice questions

##### Question 1

Given that $\frac{dy}{dx}=4x+7$dydx=4x+7, consider the following.

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 $\left(3,41\right)$(3,41).

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

##### Question 2

Consider the expression $\frac{dy}{dx}=\sqrt{6x+66}$dydx=6x+66.

1. Find an equation for $y$y in terms of $x$x, giving your answer in surd form.

Use $C$C to represent any constant.

2. Find the value of $C$C if the curve given by the equation $y=\frac{1}{9}\sqrt{\left(6x+66\right)^3}+C$y=19(6x+66)3+C passes through the point $\left(-5,17\right)$(5,17).

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

## Applications of anti-derivatives

Anti-derivatives can be applied in many contexts such as business, motion and science. The key to solving these problems is determining the primitive function and interpreting the conditions provided to work out the associated constant value.

### Motion of a body

We have seen that velocity is the rate of change of displacement with respect to time and acceleration is the rate of change of velocity. So, given velocity as a function of time, we obtain displacement as a function of time by anti-differentiation. Similarly, a velocity function can be obtained from an acceleration function by anti-differentiation.

The integral sign is used for anti-differentiation. Thus for displacement $x$x, velocity $v$v and acceleration $a$a, we have:

$x(t)=\int\ v(t)\ dt+C$x(t)= v(t) dt+C

$v(t)=\int\ a(t)\ dt+C$v(t)= a(t) dt+C

#### Worked examples

##### Example 2

Consider a body moving so that its velocity in 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, find its displacement function $x(t)$x(t). What is the displacement at $t=3$t=3?

Think: Since we are given a velocity function are asked to find displacement, we can use  $x(t)=\int\ v(t)\ dt+C$x(t)= v(t) dt+C.

Do: Integrating the velocity function we get:

 $x(t)$x(t) $=$= $\int5t^2-2t+3\ dt$∫5t2−2t+3 dt $x(t)$x(t) $=$= $\frac{5t^3}{3}-t^2+3t+C$5t33​−t2+3t+C

We have been given the condition that at time $0$0 our displacement is $-2$2 and hence $x(0)=-2$x(0)=2. This allows us to narrow down our family of primitives to one as follows:

 $-2$−2 $=$= $\frac{5(0)^3}{3}-(0)^2+3(0)+C$5(0)33​−(0)2+3(0)+C $C$C $=$= $-2$−2

Hence, the displacement function is $x(t)=\frac{5t^3}{3}-t^2+3t-2$x(t)=5t33t2+3t2.

To find the displacement at $t=3$t=3, we use our displacement function and find $x(3)$x(3):

 $x(3)$x(3) $=$= $\frac{5\times27}{3}-9+9-2$5×273​−9+9−2 $=$= $43$43

Therefore, the displacement at $t=3$t=3 seconds is $43$43 m.

##### Example 3

A falling body near the surface of the earth has a constant acceleration of about $-9.8$9.8 m/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 $200$200 m, find its displacement function.

Think: Acceleration is the time rate of change of velocity. So we can use $v(t)=\int a(t)\ dt+C$v(t)=a(t) dt+C:

Do:

 $v(t)$v(t) $=$= $\int(-9.8)\ dt$∫(−9.8) dt $v(t)$v(t) $=$= $-9.8t+C$−9.8t+C

We are told that the body is at rest at $t=0$t=0 hence, $v(0)=0$v(0)=0. Substituting into our velocity function gives $C=0$C=0.

We can use  $x(t)=\int\ v(t)\ dt+C$x(t)= v(t) dt+C to find our displacement function:

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

At $t=0$t=0, $x(0)=200$x(0)=200. So, $C=200$C=200 and the displacement function is $x(t)=-4.9t^2+200$x(t)=4.9t2+200.

#### Practice questions

##### Question 3

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 4

The total revenue, $R$R (in thousands of dollars), from producing and selling a new product, $t$t weeks after its launch, is such that $\frac{dR}{dt}=401+\frac{500}{\left(t+1\right)^3}$dRdt=401+500(t+1)3.

1. Determine the revenue function $R$R in terms of $t$t, expressing your answer in positive index form.

Use $C$C as the constant of integration.

2. Given that the initial revenue at the time of launch was $0$0, solve for the constant $C$C and hence state the revenue function.

3. Determine the average revenue earned over the first $5$5 weeks. Round your answer to 2 decimal places.

4. Determine the revenue earned in the $6$6th week.

### Outcomes

#### 0606C14.8

Integrate sums of terms in powers of x including 1/x and 1/(ax +b).