(IC1) Particular Antiderivatives#

By the end of the lesson you will be able to:

  • Find the particular antiderivative of a function given an initial condition.

  • Given \(s(t)\), \(v(t)\), or \(a(t)\) calculate \(s(t)\), \(v(t)\), or \(a(t)\) using either differentiation or integration as needed.

Lecture Videos#

Particular Antiderivatives#

When we find the general antiderivative of a function, we are actually finding all antiderivatives up to a constant. This is why we put a \(+C\) at the end: there are actually an infinite amount of antiderivatives, all of which differ by a constant.

However, if we know a little more information about the original function we’re trying to recover, we can actually find a value for \(+C\).

We sometimes refer to this additional information as an initial condition.

Example 1#

Find the particular antiderivative \(f\), of the function:

\[ f'(x)=\dfrac{x+1}{\sqrt{x}} \]

such that \(f(1)=5\).

Click through the tabs to see the steps of our solution.

Note that when we calculate an indefinite integral, we need to:

  • remember to include the \(+C\) at the end.

We start by finding the antiderivative of \(f'(x)\) to get the general antiderivative \(f\):

\[ f(x)= \int f'(x)\; dx = \int \dfrac{x+1}{\sqrt{x}} \; dx \]

To calculate this integral we need to first do the division to get this into a form we can integrate (the sum of two power functions):

\[\begin{split} \begin{aligned} f(x) = \int \dfrac{x+1}{\sqrt{x}} \; dx & = \int \left( \dfrac{x}{\sqrt{x}} + \dfrac{1}{\sqrt{x}} \right)\; dx\\ & \\ & = \int \left( \sqrt{x} + \dfrac{1}{\sqrt{x}} \right)\; dx\\ & \\ & = \int \left( x^{1/2} + x^{-1/2} \right)\; dx\\ & \\ & = \tfrac{2}{3} x^{3/2} + \tfrac{2}{1}x^{1/2} + C\\ & \\ & = \tfrac{2}{3} x^{3/2} + 2x^{1/2} + C\\ \end{aligned} \end{split}\]

Our next step is to find the value our arbitrary constant of integration \(C\) using the initial condition: \(f(1)=5\).

This initial condition is telling us that when we plug \(x=1\) into our antiderivative we should get \(5\) out.

What we have
\[\begin{split} \begin{aligned} f(1) & = \tfrac{2}{3} \cdot 1^{3/2} + 2 \cdot 1^{1/2} + C \\ & = \dfrac{2}{3} + 2 + C \\ & = \dfrac{8}{3} + C \\ \end{aligned} \end{split}\]
What we want
\[\begin{split} \begin{aligned} f(1) & = 5 \\ \end{aligned} \end{split}\]

Setting these equal to each other and solving for \(C\) we get:

\[ \dfrac{8}{3} + C = 5 \longrightarrow C = \dfrac{7}{3} \]

Finally, we take our antiderivative and plug-in the value we found for \(C\).

\[ f(x) = \tfrac{2}{3} x^{3/2} + 2x^{1/2} + \dfrac{7}{3} \]

Example 2#

Find the particular antiderivative \(f\), of the function:

\[ f''(x)=12x^2+6x-4 \]

such that \(f(0)=4\) and \(f(1)=1\).

Click through the tabs to see the steps of our solution.

Note that when we calculate an indefinite integral, we need to:

  • remember to include the \(+C\) at the end.

We start by finding the antiderivative of \(f''(x)\) to get \(f'(x)\):

\[\begin{split} \begin{aligned} f'(x) &= \int f''(x)\; dx & = \int 12x^2+6x-4 \; dx \\ & = 4x^3+3x^2-4x+C \\ \end{aligned} \end{split}\]

Next we find the antiderivative of \(f(x)\) to get \(f(x)\):

\[\begin{split} \begin{aligned} f(x) &= \int f'(x)\; dx & = \int 4x^3+3x^2-4x+C \; dx \\ & = x^4+x^3-2x^2+Cx +D \\ \end{aligned} \end{split}\]

Our next step is to find the values our arbitrary constant of integration \(C\) using the initial condition: \(f(1)=5\).

This initial condition is telling us that when we plug \(x=1\) into our antiderivative we should get \(5\) out.

What we have
\[\begin{split} \begin{aligned} f(1) & = \tfrac{2}{3} \cdot 1^{3/2} + 2 \cdot 1^{1/2} + C \\ & = \dfrac{2}{3} + 2 + C \\ & = \dfrac{8}{3} + C \\ \end{aligned} \end{split}\]
What we want
\[\begin{split} \begin{aligned} f(1) & = 5 \\ \end{aligned} \end{split}\]

Setting these equal to each other and solving for \(C\) we get:

\[ \dfrac{8}{3} + C = 5 \longrightarrow C = \dfrac{7}{3} \]

Motion in a Straight Line#

When we are discussing an object moving in a straight line there are 3 important functions to consider:

  • Position Function \(s(t)\) - Location of the object at time \(t\).

  • Velocity Function \(v(t)\) - Speed and direction of the object at time \(t\).

  • Acceleration Function \(a(t)\) - Describes whether the object is speeding up, slowing down, or changing direction at time \(t\).

These three functions are all related by differentiation:

\[ s(t) \xrightarrow{\text{Differentiation}} v(t)=s'(t) \xrightarrow{\text{Differentiation}} a(t)=v'(t)= s''(t) \]

and by integration:

\[ a(t) \xrightarrow{\text{Integration}} v(t)=\int a(t) \, dt \xrightarrow{\text{Integration}} s(t)=\int v(t) \, dt \]

Example 3#

A rocket is fired vertically into the air. The height (position) of the rocket is the distance measured from the top of the rocket to the launch pad. Initially, the top of the rocket is \(8\) meters above the launch pad. The velocity of the rocket \(t\) seconds after liftoff is:

\[ v(t)= 6t+0.5 \]

Find the height of the rocket at time \(t\).

Example 4#

A ball is thrown upward with a speed of 48 ft/s from the edge of a cliff \(432\) ft above the ground.

  1. Find the height of the ball above the ground after \(t\) seconds.

  2. When does the ball reach its maximum height?

  3. When does it hit the ground?