What is the average distance between a point in the interval [-4,3] and the origin?

Question

Cameron Alexander · Accepted Answer

I get #25/14# The distance between a point #x# on the number line and the origin is #absx# We have been asked for the average value of this function on the interval #[-4,3]# We can either evaluate #1/(3-(-4)) int_-4^3 absx dx# or work through the equivalent geometry without the integral. The area under graph of #absx# and above #[=4,3]# consists of two isosceles right triangles. On the left we have a triangle with base and height #4#, so its area is #8#. On the right we have a triangle with base and height #3#, so its area is #9/2#. The total area under the graph is #25/2# Divide by the total base of #3+4 = 7# to get an average value of #25/14#

Ezra Adams · Answer

undefined To find the average distance between a point in the interval [-4, 3] and the origin, we need to calculate the average value of the distance function over that interval. The distance between any point (x, 0) and the origin (0, 0) in the coordinate plane is given by the absolute value of x. Therefore, the distance function in this case is |x|.

To find the average distance, we integrate |x| over the interval [-4, 3] and then divide by the length of the interval, which is 7 (3 - (-4)).

$$ \text{Average distance} = \frac{1}{7} \int_{-4}^{3} |x| \, dx $$

$$ = \frac{1}{7} \left[ \int_{-4}^{0} -x \, dx + \int_{0}^{3} x \, dx \right] $$

$$ = \frac{1}{7} \left[ \left[ -\frac{x^2}{2} \right]_{-4}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{3} \right] $$

$$ = \frac{1}{7} \left[ \left( 0 - \left( -\frac{(-4)^2}{2} \right) \right) + \left( \frac{3^2}{2} - 0 \right) \right] $$

$$ = \frac{1}{7} \left[ \left( 0 - 8 \right) + \left( \frac{9}{2} \right) \right] $$

$$ = \frac{1}{7} \left( -8 + \frac{9}{2} \right) $$

$$ = \frac{1}{7} \left( -\frac{7}{2} \right) $$

$$ = -\frac{1}{2} $$

So, the average distance between a point in the interval [-4, 3] and the origin is $ \frac{1}{2} $.