00:01
In this problem we are given that the marginal cost mc of producing x units is 9 over square root of x and we are asked to find out the total cost c of x of producing the first 64 units, that is from x equals to 0 up to x equals to 64.
00:29
The marginal cost is the first derivative of the cost function.
00:38
So in order to find out the cost function c of x, we must integrate the marginal cost with respect to x.
00:46
So we get c of x to be equal to the integral of 9 over square root of x d x with the limits of integration being 0 up to 64 since we need to calculate the cost of producing the first 64 units.
01:05
So we have integral 0 to 64 here.
01:08
The integrant can be rewritten.
01:11
We have 9 times integral 0 to 64 of x par negative 1 over 2 dx.
01:20
The antiderivative of x par n is x par n plus 1 the whole divided by n plus 1.
01:28
In this case we have x par negative 1 over 2.
01:31
So we get x par negative 1 over 2 plus 1 which is positive 1 over 2 the whole divided by 1 over 2.
01:39
So let us substitute this in the integral...