00:01
So in this question, they say, i would find the arc length of the graph of this function over the indicated interval.
00:07
And i have here, y equals two thirds times x of the three has power plus one.
00:13
So how do we find the arc length of a function over a given interval? so we have a formula, and my formula is that arc length is equal to the integral from a to b of the square root of 1 plus.
00:32
F prime of x being squared all of this d x so this time what are my limits of integration well i appear to be on the x interval this time starting at x equals zero and ending at x equals two and so i'm going to have limits of integration this time from zero to two now i'm going to have the square root of one plus i need my prime of x being square.
01:07
So we're going to need our derivative.
01:11
So what is y prime this time? well, if i take the derivative of two -thirds, x to the three -half power, two -thirds times three -haves is one, x to the one -half power.
01:26
Of course, that plus one at the end, that differentiates away to 0 so that i'm getting my f prime of x is x to the 1 half power again that is being squared all of this d x now if i square x to the 1 half power that's just x so that i'm getting an interval from 0 to 2 of the square of 1 plus x d x now in order to evaluate this definite interval i'm going to find an anti -derivitant.
02:06
To do that, remember that the square root of 1 plus x that can be thought of as the quantity of 1 plus x being raised to the 1 -half power.
02:17
So my anti -derivative, i'm going to add 1 to the exponent, 1 plus x to the 3f, and divide by that new exponent.
02:26
Dividing by 3 halves, i multiply by 2 thirds...