Why do we need a plus C after indefinite integrals?

For example, think of the following function

$$\begin{align} f(x) = 2x + 4 \end{align}$$

We can take the derivative of this function

$$\begin{align} f'(x) = 2 \end{align}$$

If we take the anti-derivative of this function

$$\begin{align} F(x) = 2x \end{align}$$

We took the derivative, then we took the anti-derivative, but we don't end up back at the original function

This is because when we took the derivative, the constant term (the plus 4) gets turned into 0

When you take the anti-derivative of 0, it becomes 0, because $\frac{d}{dx} 0 = 0$

This happens with any constant term being added, the derivative of it turns to 0 and the anti-derivative can't get it back

This is why we need the plus C, it represents an arbitrary constant to account for all possible anti-derivatives

The plus C stands for the Constant Of Integration

This means the indefinite integral is just an anti-derivative with the constant of integration

$$\begin{align} \int{f(x) \ dx} = F(x) + C \end{align}$$

Also, the integral of 0 is $C$

$$\begin{align} \int{0 \ dx} = C \end{align}$$

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Another Example

If you try to find the following indefinite integral

$$\begin{align} \int{2 \sin(x) \cos(x) \ dx} \end{align}$$

There are 2 ways you can solve this, by using trig properties or U-sub

$$\begin{align} \int{\sin(2x) \ dx} = -\frac{1}{2}\cos(2x) + C \end{align}$$

$$\begin{align} \int{2u \ du} = 2\frac{u^2}{2} + C = \sin^2(x) + C &&\text{$u = \sin(x)$} \end{align}$$

These 2 anti-derivatives are different, but they have the same derivative, this is because of the $+ C$

If you solve for $C$ you get a value of $\pm \frac{1}{2}$

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Combining Multiple $+C$

A lot of the time, you solve multiple integrals that are being added or subtracted

Each of these integrals results in it's own $+C$ term

But since each of the $C$ are constant, and they are only being added or subtracted, this means you can combine them together

This is because a constant plus another constant is also a constant

For example:

$$\begin{align} \int{f(x) \ dx} + \int{g(x) \ dx} + \int{h(x) \ dx} &= F(x) + C_1 + G(x) + C_2 + H(x) + C_3 \\ &= F(x) + G(x) + H(x) + C \end{align}$$

Because of this, I like to wait until the end to account for the $+C$

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Finding The Value of $C$

Even though $C$ is an arbitrary constant, we can find the value of it given a point on the original function

For example

Given:

$$\begin{align} f'(x) &= 4x \\ f(2) &= 10 \\ \end{align}$$

Find the Original Function $f(x)$

First, we need to find the integral in order to get from the derivative to the original function

$$\begin{align} f(x) = \int{f'(x) \ dx} &= \int{4x} = 2x^2 + C \end{align}$$

We can plug 1 in for $x$ and set it equal to 8 to find C

$$\begin{align} 2(2)^2 + C &= 10 \\ 8 + C &= 10 \\ C &= 2 \\ \end{align}$$

We can combine this with our anti-derivative and get:

$$\begin{align} f(x) = 2x^2 + 2 \end{align}$$

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Practice Problems

• Find $f(x)$ if $f'(x) = cos(x)$ and $f(\pi) = 3$

  $$\begin{align} f(x) = \int{cos(x) \ dx} = sin(x) + C \end{align}$$

  Now we need to find the value of $C$

  $$\begin{align} sin(\pi) + C &= 3 \\ 0 + C &= 3 \\ C &= 3 \\ \end{align}$$

  Plug this back into the integral for the original function

  $$\begin{align} f(x) = sin(x) + 3 \end{align}$$

• Find $f(x)$ if $f'(x) = e^x$ and $f(0) = 0$

  $$\begin{align} f(x) = \int{e^x \ dx} = e^x + C \end{align}$$

  Now we need to find the value of $C$

  $$\begin{align} e^{0} + C &= 0 \\ 1 + C &= 0 \\ C &= -1 \\ \end{align}$$

  Plug this back into the integral for the original function

  $$\begin{align} f(x) = e^x - 1 \end{align}$$

• Find $f(x)$ if $f''(x) = 3x$ and $f'(4) = 0$ and $f(1) = 2$

  Here we need to apply the previous processes twice

  $$\begin{align} f'(x) = \int{3x \ dx} = \frac{3x^2}{2} + C_1 \end{align}$$

  Now we need to find the value of $C_1$

  $$\begin{align} \frac{3(4)^2}{2} + C_1 &= 0 \\ 24 + C_1 &= 0 \\ C_1 &= -24 \\ \end{align}$$

  Plug this back into the integral for $f'(x)$

  $$\begin{align} f'(x) = \frac{3x^2}{2} - 24 \end{align}$$

  Now we need to repeat this again

  $$\begin{align} f(x) = \int{\left(\frac{3x^2}{2} - 24\right) \ dx} = \frac{x^3}{2} - 24x + C_2 \end{align}$$

  Now we need to find the value of $C_2$

  $$\begin{align} \frac{(1)^3}{2} - 24(1) + C_2 &= 2 \\ \frac{1}{2} - 24 + C_2 &= 2 \\ C_2 &= 25.5 \\ \end{align}$$

  Plug this back into the integral for $f(x)$

  $$\begin{align} f(x) = \frac{x^3}{2} - 24x + 25.5 \end{align}$$