00:01
So in this question, we are trying to find the area that lies beneath the graph of the quantity y equals 3 minus x times the squared of x.
00:10
So we have this function, y equals the quantity of 3 minus x times the square of x.
00:18
We're trying to find the area that lies beneath that curve and above the x -axis from x equals 0 to x -equals 3.
00:27
So to do that, we're going to set up a definite integral and where does my region start? well, my region starts at x equals zero, right? so my region starts at x equals zero, and it ends at x equals three.
00:43
So i'm going to have an integral from zero to three of this function of the quantity of three minus x times the square root of x, dx, right? and so how am i going to evaluate this definite integral? well, i'm going to need an antiderivis.
01:02
Yes but before i get my anti -derivative how can i make that process easier well i could distribute that square root x and i could say this is the integral from zero to three of the quantity of three root x minus x root x dx yes now what do i know about the square of x.
01:31
I know that it's x to the one half power.
01:34
So i could say this is the integral from zero to three of the quantity of three x to the one half power minus x times x to the one half power dx.
01:51
Now let's do this multiplication, x times x to the one half.
01:57
That's really x to the first times x to the one half.
02:02
What do i do to those exponent? well, i add them, right? and so that second term here, that second term, x to the first, times x to the one -half, that's becoming x to the three -haves, right? i add those exponents, one plus one -half, to get three - halves.
02:28
And so now i'm ready to find my anti -derivative.
02:33
So how am i going to get that anti -derivative? well, i'm going to get that anti -derivative.
02:37
Well, i'm going to going to start by, in that first term, adding one to the exponent and dividing by the new exponent.
02:45
So 3x to the three halves, divided by three halves.
02:51
And for that second term, same idea...