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Numerade Educator



Problem 13 Easy Difficulty

If the amount of capital that a company has at time $ t $ is $ f(t) $, then the derivative $ f^{\prime} (t) $, is called the net investment flow. Suppose that the net investment flow is $ \sqrt{x} $ million dollars per year (where $ t $ is measured in years). Find the increase in capital (the capital formation) from the fourth year to the eighth year.


$$\frac{2}{3}(16 \sqrt{2}-8) \approx \$ 9.75 \text { million }$$


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Video Transcript

all right. So the question gives us a net investment flow function of square of tea, which it's Dr as being F prime of tea, where FF t is the amount of capital and then asked us to find the increase in capital from the fourth year to the eighth year. So the quantity that were interested in then is F of eight, which is the quantity of capital minus half of four, which is also the quantity of capital for the 1st 1 It's at your eight. Second one is that your four? Their difference will be the increase and so on the side we see than that change hero, which tells us that we can integrate from a to b the derivative and get the difference in values. So our quantity that we're interested in is just the integral from 4 to 8 of the square root of tea duty. So evaluating that is going to be to over three T 23 halves from 4 to 8. There's equal to two or three when you plug in eight. There we have the square root of 5 12 than minus new plug and four you get squared of 64 and that is going to be equal to to over three times the quantity. 16 route, too minus state. Yeah, approximately 9.75 $1,000,000. And that right there, since it's what are integral is button that changed here? Um, it will be the difference of f of eight in F four. So 9.75 million is the increase in capital.