# What is the antiderivative of f(x) = x - 7?

Mar 19, 2016

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

#### Explanation:

The antiderivative of $f \left(x\right) = x - 7$ is quite simple. It just involves two cases of the reverse power rule. First, we write down the problem in math notation:
$\int x - 7 \mathrm{dx}$

The sum rule for integrals (antiderivatives) says we can simplify this to:
$\int x \mathrm{dx} - \int 7 \mathrm{dx}$

Furthermore, another rule of integrals says that $\int a x \mathrm{dx} = a \int x \mathrm{dx}$; in other words, we can pull constants out of the integral. Since the $7$ in the second integral is constant, we can say:
$\int x \mathrm{dx} - 7 \int \mathrm{dx}$

Solving these is easy. The power rule for derivatives says that to find the derivative of ${x}^{a}$, multiply $x$ by $a$ and then decrease $a$ by one (such as $\frac{d}{\mathrm{dx}} \left({x}^{2}\right) = 2 \cdot {x}^{2 - 1} = 2 {x}^{1} = 2 x$). It makes sense, then, that the antiderivative would be this in reverse: increase $a$ by one, then divide by the new power (so the antiderivative of $2 x$ would be $2 {x}^{1 + 1} = 2 {x}^{2} = \frac{2 {x}^{2}}{2} = {x}^{2}$, which is the original function). Applying this to our problem:
$\int x \mathrm{dx} = {x}^{2} / 2 + {C}_{1}$ and $- 7 \int \mathrm{dx} = - 7 x - {C}_{2} \to$ this one is the so-called "perfect integral"

Our solution is therefore:
${x}^{2} / 2 + {C}_{1} - 7 x - {C}_{2}$

And since ${C}_{1} - {C}_{2}$ is just another constant, we can combine them into a general constant $C$ to get our final result of:
${x}^{2} / 2 + 7 x + C$