Constant of Integration (College Board AP® Calculus AB)

Study Guide

Test yourself

Written by: Roger B

Reviewed by: Dan Finlay

Finding the constant of integration

How can I find the value of a constant of integration?

When finding an indefinite integral, a constant of integration is needed
- $\int f (x) d x = F (x) + C$
  - where $F^{'} (x) = f (x)$
  - and $C$ is any constant
If you know more information about $F (x)$ you can work out the value of the constant $C$
- You may be given the value of $F (x)$ for some particular value of $x$
- Or you may be told that the graph of $F (x)$ goes through a particular point $(x_{0}, y_{0})$
  - Remember in this case that $y_{0} = F (x_{0})$
- This lets you set up and solve an equation to find the value of $C$
For example, if the function $F$ is defined by $F (x) = \int (2 x + 3) d x$ , and you also know that the graph of $F$ goes through the point $(1, 2)$
- First integrate
  - $F (x) = \int (2 x + 3) d x = x^{2} + 3 x + C$
- The graph goes through $(1, 2)$ , so $F (1) = 2$
  - $F (1) = {(1)}^{2} + 3 (1) + C = 2$
- Solve the equation for $C$
  - $\begin{array}{rcl} 1 + 3 + C & = & 2 \Rightarrow C + 4 = 2 \Rightarrow C = - 2 \end{array}$
- Therefore
  - $F (x) = x^{2} + 3 x - 2$
Finding the constant of integration in this way is equivalent to finding the particular solution of a first-order differential equation in the form $\frac{d y}{d x} = f (x)$
- See the 'Particular Solutions' study guide for a more formal treatment of this

Worked Example

The function $h$ is being used to measure the height of a projectile above the ground at time $t$ . The height $h (t)$ is measured in feet, and $t$ is measured in seconds.

It is known that $h$ satisfies the equation $h (t) = \int (70 - 32 t) d t$ , and that at time $t = 2$ the projectile is 81 feet above the ground.

Find an explicit expression for $h$ in terms of $t$ .

Answer:

First find the indefinite integral; don't forget the constant of integration

$h (t) = \int (70 - 32 t) d t = 70 t - 16 t^{2} + C$

We know that $h (2) = 81$

$70 (2) - 16 {(2)}^{2} + C = 81$

Solve for $C$

$\begin{array}{rcl} 140 - 64 + C & = & 81 \\ C + 76 & = & 81 \\ C & = & 5 \end{array}$

Substitute that value for $C$ into the expression for $h (t)$

$h (t) = 70 t - 16 t^{2} + 5$

Last updated: 15 August 2024

You've read 0 of your 5 free study guides this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Test yourself

Did this page help you?

Previous:Indefinite Integral RulesNext:Accumulation of Change