# Indefinite Integration

• Feb 11th 2009, 11:12 AM
Macleef
Indefinite Integration
Let f(x) represent the total number of items a subject has memorized x minutes after being presented with a long list of items to learn. Psychologists refer to the graph of y = f(x) as a learning curve and to f'(x) as the learning rate. The time of peak efficiency is the time when the learning rate is maximized. Suppose the learning rate is

$f'(x) = 0.1 (10 + 12x - 0.6x^2)$ for $0 \leq x \leq 25$

a) When does peak efficiency occur? What is the learning rate of peak efficiency?

b) What is f(x)?

a) Find f(0), f(25)

$0 = 0.1(10 + 12x - 0.6x^2)$

$10 + 12x - 0.6x^2 = 0$

$12x - 0.6x^2 = -10$

$x = 0$

$1 - 200x = 0$

$1 = 200x$

$\frac{1}{200} = x$

...but the answer is x = 10 and I don't even know where this number comes from! Plus, the critical points I found are outside the restrictions...

b) $\int 1 \, dx + \int 1.2x \, dx - \int 0.06x^2 \, dx$

$= x + 0.6x^2 - 0.02x^3 + c$

...my question is why is there no "c" (arbitrary constant)??

=====================

A manufacturer estimates marginal revenue to be $200q^{-\frac{1}{2}}$ dollars per unit when the level of production is q units. The corresponding marginal cost has been found to be 0.4q dollars per unit. If the manufacturer's profit is $2,000 when the level of population is 25 units, what is the profit when the level of production is 36 units? $\int 200q^{-\frac{1}{2}} + 0.4q\, dq$ $= 400q^{\frac{1}{2}} + 0.2q^2 + c$ $2000 = 400(25)^{\frac{1}{2}} + 0.2(25)^2 + c$ $2000 = 2125 + c$ $-125 = c$ $R(q) = 400q^{\frac{1}{2}} + 0.2q^2 - 125$ $R(36) = 2534.20$ ...the right answer is$2265.80, what am I doing wrong?
• Feb 11th 2009, 12:49 PM
Rincewind
Quote:

Originally Posted by Macleef
Let f(x) represent the total number of items a subject has memorized x minutes after being presented with a long list of items to learn. Psychologists refer to the graph of y = f(x) as a learning curve and to f'(x) as the learning rate. The time of peak efficiency is the time when the learning rate is maximized. Suppose the learning rate is

$f'(x) = 0.1 (10 + 12x - 0.6x^2)$ for $0 \leq x \leq 25$

a) When does peak efficiency occur? What is the learning rate of peak efficiency?

b) What is f(x)?

a) Find f(0), f(25)

$0 = 0.1(10 + 12x - 0.6x^2)$

$10 + 12x - 0.6x^2 = 0$

$12x - 0.6x^2 = -10$

$x = 0$

$1 - 200x = 0$

$1 = 200x$

$\frac{1}{200} = x$

...but the answer is x = 10 and I don't even know where this number comes from! Plus, the critical points I found are outside the restrictions...

You need to find the maximum in $f'(x)$ so what you need to calculate is the value of x for which $f''(x) = 0$

$f'(x) = 0.1 (10 + 12x - 0.6x^2)$

$f''(x) = 0.1 (12 - 1.2x)$

So solving for $f''(x) = 0$ you get $x=10$ so the only critical point occurs for this value. Now determine if it is a maximum or minimum you need to look at the next derivative, i.e. $f'''(x)$ you will find the sign is negative and therefore x=10 is a global maximum.

The rate of peak efficiency is just

$f'(10) = 0.1 (10 + 12\times 10 - 0.6\times 10^2)$

Quote:

Originally Posted by Macleef
b) $\int 1 \, dx + \int 1.2x \, dx - \int 0.06x^2 \, dx$

$= x + 0.6x^2 - 0.02x^3 + c$

...my question is why is there no "c" (arbitrary constant)??

One may assume that at time $x=0$ the subject has not memorised anything so

$f(0) = 0$

This can be used to eliminate the constant. ( $c=0$).

Quote:

Originally Posted by Macleef
A manufacturer estimates marginal revenue to be $200q^{-\frac{1}{2}}$ dollars per unit when the level of production is q units. The corresponding marginal cost has been found to be 0.4q dollars per unit. If the manufacturer's profit is $2,000 when the level of population is 25 units, what is the profit when the level of production is 36 units? $\int 200q^{-\frac{1}{2}} + 0.4q\, dq$ $= 400q^{\frac{1}{2}} + 0.2q^2 + c$ $2000 = 400(25)^{\frac{1}{2}} + 0.2(25)^2 + c$ $2000 = 2125 + c$ $-125 = c$ $R(q) = 400q^{\frac{1}{2}} + 0.2q^2 - 125$ $R(36) = 2534.20$ ...the right answer is$2265.80, what am I doing wrong?

The only thing you have done wrong is to have the wrong sign for the marginal cost. As this is a cost term it should be negative. So start with

$\int 200q^{-\frac{1}{2}} - 0.4q\, dq$

and you should get the correct answer.

Hope this helps.