# How do you find the most general antiderivative of the function for f(x) = x - 7?

Oct 20, 2015

${x}^{2} / 2 - 7 x + C$

#### Explanation:

The general antiderivative of $f \left(x\right)$ is $F \left(x\right) + C$, where $F$ is a differentiable function. All that means is that if you differentiate the antiderivative, you get the original function - so to find the antiderivative, you reverse the process of finding a derivative.

Sound confusing? Easier done than said. What we're doing is only taking the indefinite integral of $f \left(x\right)$ - in other words, $\int x - 7 \mathrm{dx}$. The properties of integrals say that we can break it up in pieces in cases of addition and subtraction; thus,

$\int x - 7 \mathrm{dx} = \int x \mathrm{dx} - \int 7 \mathrm{dx}$.

Further using the properties of integrals,

$\int x - 7 \mathrm{dx} = \int x \mathrm{dx} - 7 \int \mathrm{dx}$

First, let's do $\int x \mathrm{dx}$. What we're asking ourselves is: what function, when you take its derivative, equals $x$? Well, ${x}^{2} / 2$, of course! Using the power rule, we multiply the expression by the exponent and then reduce the exponent by one; doing that gives $2 \cdot \frac{{x}^{2 - 1}}{2} = x$. So, our first integral reduces to ${x}^{2} / 2 + C$.

Now, why the $C$? We put the $C$ (which is just a constant - any old number, like $2$, $\sqrt{5}$, and $\pi$) because we're finding the general antiderivative. Thus, we don't know if there's another number hiding in our antiderivative - so we put the $C$ there to make it general and cover our behinds.

Finally, we evaluate $7 \int \mathrm{dx}$. This ($\int \mathrm{dx}$) is called a perfect integral because its result is plain ol' $x$. Since we have a $7$ in front of it, our final result is $7 x + C$ (never forget the $C$!).

We can finally put our pieces together for the final answer:

$\int x - 7 \mathrm{dx} = \int x \mathrm{dx} - \int 7 \mathrm{dx}$
intx-7dx = (x^2/2 + C) - (7x + C)
$= {x}^{2} / 2 + C - 7 x - C$ (distributing the negative sign)

You might think $C - C = 0$, but that's not quite right. Recall that $C$ is any number - both of them. So one $C$ can be $4$ and the other can be $3$, in which case $C - C = 1$ or $- 1$. But then again, $1$ and $- 1$ are constants, right? In fact, $C - C$ will always be a constant, and since $C$ represents a constant, we can just call $C - C$ normal $C$. Take my word for it.

Thus, the final result is ${x}^{2} / 2 - 7 x + C$.