00:01
Problem we're given a marginal cost function for producing our product, which is 74 plus 1 .1x minus 2 times 10 to the negative 3 x squared plus 4 times 10 to the negative 5 x cube.
00:15
And we have that we need to find the change in cost going from 1 ,200 to 1 ,600 units, which we write as just the cost of 1 ,600 units minus the cost of 1 ,600 units 1 ,200 units.
00:34
And we have the following relation, which we should know, which is that the derivative of cost with respect to x is equal to the marginal cost of x.
00:48
So we're going to first take this relation, take this relation that we have here and use it.
01:06
But we're going to plug in for marginal cost of x with the marginal cost function that they go.
01:12
Gave us.
01:18
So that's and there's not supposed to be a point there.
01:33
All right.
01:41
So we've plugged in and we want to find the cost of 1600 units minus the cost of 1 ,200 units.
01:49
And if you remember how we evaluate an integral, a definite integral, some integral of f of x from a to b, dx, that's just equal to the anti -derivative of f of b minus the anti -derivative of f of a which you'll notice looks exactly like what we want to find so what we're going to do is integrate with bounds a and b a and b so that um b so that we get f of 1600 minus f of 1200 and in this case our function is actually c not f but we choose our bounds so that we have 1600 here plugged in for b b equals 1600 and our lower bound is going to be 1200 just like in our solution we want to find up here so that's where we go from here once i can get this erased so on both sides we're integrating not erasing and integrating dx and our lower bound we said it was going to be 1200 and our upper bound we said was going to be 1600 and we have to do that on both sides to keep the equal sign and dx and so the left side of our equal sign it's easy to evaluate we just integrate and plug in our upper and lower bounds so we get c of sixteen hundred because we're integrating the derivative of c which just gives us c minus c of 1200 and then equals now we need to use the power rule, which says that the integral of x to the a, dx, equals x to the a plus 1 divided by a plus 1.
04:23
So, and we need to do that for each term of our polynomial...