# How do you find the average distance from the origin of a point on the parabola y=x^2, 0<=x<=4 with respect to x?

Jan 1, 2018

$\frac{17 \sqrt{17} - 1}{12} \approx 5.84$

#### Explanation:

Let $P$ be a point on the parabola. The coordinates with respect to $x$ will be $\left(x , {x}^{2}\right)$.

We can make a function out of the distance to the origin using the distance formula:
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

Applying this to our point $P = \left(x , {x}^{2}\right)$ and the origin $\left(0 , 0\right)$, we get:
$d \left(x\right) = \sqrt{{\left(x - 0\right)}^{2} + {\left({x}^{2} - 0\right)}^{2}}$

$d \left(x\right) = \sqrt{{x}^{2} + {x}^{4}}$

To work out the average value of the function, we can use the average value formula. For a function $f \left(x\right)$ on the interval $\left[a , b\right]$, the average value will be:
$\frac{{\int}_{a}^{b} f \left(x\right) \setminus \mathrm{dx}}{b - a}$

In our case, we get:
$\frac{{\int}_{0}^{4} d \left(x\right) \setminus \mathrm{dx}}{4 - 0} = \frac{1}{4} {\int}_{0}^{4} \sqrt{{x}^{2} + {x}^{4}} \setminus \mathrm{dx}$

I will first work out the antiderivative:
$\int \setminus \sqrt{{x}^{2} + {x}^{4}} \setminus \mathrm{dx} =$

=int\ sqrt(x^2(1+x^2)\ dx=

$= \int \setminus x \sqrt{{x}^{2} + 1} \setminus \mathrm{dx} =$

Now we can introduce a u-substitution with $u = {x}^{2} + 1$. The derivative of $u$ is then $2 x$, so we divide through by that:
$= \int \setminus \frac{\cancel{x} \sqrt{u}}{2 \cancel{x}} \setminus \mathrm{du} =$

$\frac{1}{2} \int \setminus \sqrt{u} \setminus \mathrm{du} = \frac{1}{2} \cdot \frac{2}{3} {u}^{\frac{3}{2}} + C = \frac{2}{6} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}} + C$

Now we can evaluate the original expression:
$\frac{1}{4} {\int}_{0}^{4} \sqrt{{x}^{4} + {x}^{2}} \setminus \mathrm{dx} = \frac{1}{4} {\left[\frac{1}{3} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}}\right]}_{0}^{4} =$

$= \frac{1}{4} \left(\frac{1}{3} {\left(1 + 16\right)}^{\frac{3}{2}} - \frac{1}{3} {\left(1 + 0\right)}^{\frac{3}{2}}\right) =$

$= \frac{1}{4} \left(\frac{1}{3} \cdot {17}^{\frac{3}{2}} - \frac{1}{3} \cdot {1}^{\frac{3}{2}}\right) = \frac{1}{4} \left(\frac{17 \sqrt{17}}{3} - \frac{1}{3}\right) =$

$= \frac{1}{4} \cdot \frac{17 \sqrt{17} - 1}{3} = \frac{17 \sqrt{17} - 1}{12}$

So, the average distance from a point to the origin on the parabola on the interval $\left[0 , 4\right]$ is:
$\frac{17 \sqrt{17} - 1}{12} \approx 5.84$