Query 7

#$&*

course 272

10/20 10p

007. `query 7

If you had trouble with the concept and/or procedures for substitution, consider submitting all or part of the document

qa_ac2_14.htm,

which is intended as an introduction to integration by substitution.

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Question: `q5.2.36 (was 5.2.2( (previously 5.2.36 (was 5.2.34) ) integral of x^2 (1-x^3)^2 by formal substitution.

What is the integral of the given function?

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Your solution:

x^2 (1-x^3)^2

u = (1-x^3)

du/dx = -3x^2

int of x^2 = x^3 / 3

int -1/3 (u^2) du

= -1/3 (u^3 /3)

=-1/9 (u^3)

=-1/0 (1-x^3)^3

confidence rating #$&*: 3

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Given Solution:

`a If we let u = 1 - x^3 then u ' = - 3 x^2 and the x^2 in our integrand is - u ' / 3.

(1-x^3)^2 is u^2, so the integrand is - u ' / 3 * u^2 = -1/3 u^3 u ' .

So the integral is you have -1/3 u^2 du. The integral of u^2 u ' is 1/3 u^3.

Thus the integral of -1/3 u^2 u ' is -1/3 of 1/3 u^3, or -1/9 u^3.

So your integral should be -1/9 u^3 = -1/9 (1-x^3)^3.

The general antiderivative is -1/9 ( 1 - x^3)^3 + c. **

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Self-critique (if necessary):

I forgot to put it in the general formula with adding c as the constant.

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Self-critique Rating: 3

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Question: `qWhat is the derivative of your result?

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Your solution:

Chain rule to check but it should be what the problem started with x^2(1-x^3)^2

-1/9 (-3x^2) * 3(1-x^3)^2 = x^2 (1-x^3)^2

confidence rating #$&*: 3

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Given Solution:

`a The derivative of -1/9 (1-x^3)^3, using the Chain Rule, is the product of -1/9, 3(1-x^3)^2, and the derivative -3x^2 of the 'inner function' (1-x^3). Multiplying these factors we get -1/9 (-3x^2) * 3(1-x^3)^2 = x^2 (1-x^3)^2. **

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Self-critique (if necessary): ok

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Self-critique Rating: 3

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Question: `q 5.2.54 (was 5.2.4) (previously 5.2.54 (was 5.2.52)) find x | dx/dp = -400/(.02p-1)^3, x=10000 when p=100

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Your solution:

x | dx/dp = -400/(.02p-1)^3

dx = -400 (.02p-1)^-3 * dp

u = (.02p-1)

du/dp = .02

du = .02dp

dp =du /.02 = 50du

int -400 * 50u^-3 * du

int -200,000u^-3 * du

int -200,000 / u^-3

int u^-3 = u^-2 / -2 = -2u^-2

= -200,000 / 2u^-2

= 10,000 / u^-2 + c

= 10,000 / (.02p-2)^-2 + c

10000 = 10,000 / (.02(100)-1^-2) + c

0 = c

solution x = 10000 /(.02p-1)^-2

@&

int -400 * 50u^-3 * du =

-20 000 int(u^-3 du) =

-20 000 * -1/2 * u^(-2) =

10 000 u^-2 =

10 000 / u^2,

not 10 000 / u^(-2)

*@

confidence rating #$&*:

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Given Solution:

`a The equation rearranges to dx = -400 * dp * (.02 p - 1)^-3.

An antiderivative of the left-hand side could be just x.

An antiderivative of dp * (.02 p - 1)^-3 is found using u = .02 p - 1, so du = .02 dp and dp = du / .02 = 50 du. Thus the right-hand side becomes -400 * 50 u^-3 du = -20000 u^-3 du, with antiderivative 20000 / 2 * u^-2 + c = 10,000 u^-2 + c.

So we have x = 10,000 * u^-2 + c = 10,000 * (.02 p - 1)^-2 + c.

Note that dx / dp is therefore 10,000 * -2 * .02 (p-1)^-3 = -400 (.02 p - 1)^-3, consistent with the original equation.

Since x = 10,000 * (.02 p - 1)^-2 + c and x = 10,000 when p = 100 we have

10,000 = 10,000 * (.02 * 100 - 1)^2 + c

10,000 = 10,000 / 1^2 + c

10,000 = 10,000 + c so

c = 0.

The solution is therefore

x = 10,000 * (.02 p - 1)^-2 + 0 or just

x = 10,000 * (.02 p - 1)^-2.

**

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Self-critique (if necessary):

I am comfortable with this problem….while comparing answers we arrived at same result but some of our pos/neg signs were off…I reworked mine but kept getting my results as typed above.

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Self-critique Rating:

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Question: `q 5.3.4 (was 5.3.1) (previously 5.3.04 (was 5.3.04)) integral of e^(-.25 x) by Exponential Rule

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Your solution:

U = -.25x

Du/dx = -.25

Du = -.25 dx

Dx = 1/-.25 dc = -4du

E^u * -4du

=-4e^u du

=-4e^u + c

=-4e^(-.25x) + c

confidence rating #$&*:

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Given Solution:

`a Simple substitution u = -.25 x gives us du/dx = -.25 so that du = -.25 dx and dx = du / (-.25) = -4 du.

Our original integrand e^(-.25 x) dx therefore becomes e^u * (-4 du) = -4 e^u du. Our general antiderivative will be -4 e^u + c, meaning -4 e^(-.25 x) + c.

The derivative of -4 e^(-.25 x) + c is -4 ( -.25 e^-.25 x) = e^-.25 x, verifying our result.

The General Exponential Rule is equivalent to this:

u = -.25 x so du/dx = -.25. Thus the integral is of e^u / (du/dx) = e^(-.25 x) / (-1/4) = -4 e^(-.25 x). *&*&

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Self-critique (if necessary):

ok

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Self-critique Rating:3

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Question: `q 5.3.10 (was 5.3.2) (previously 5.3.10 (was 5.3.10)) integral of 3(x-4)e^(x^2-8x) by Exponential Rule

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Your solution:

3(x-4)e^(x^2-8x)

u = (x^2-8x)

du/dx = 2x-8

3*1/2(x-4)e^u du/dx

=3/2 du/dx e^u

=3/2 e^u du/dx

=3/2 e^(x^2-8x) + c

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a if u=x^2 - 8x then du / dx = 2x - 8

x-4 = 1/2(2x-8) so 3(x-4) = 3/2 du/dx.

Thus 3(x-4)e^(x^2-8x) is 3/2 e^u du/dx.

The general antiderivative of e^u du/dx is e^u + c, so the integral of 3/2 e^u du/dx is 3/2 e^u.

Substituting x^2 - 8x for u we have 3/2 e^(x^2-8x) + c. **

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `qproblem 5.3.16 (was 5.3.3) (previously 5.3.16) integral of 1/(6x-5) by Log Rule

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Your solution:

1/(6x-5)

int 1/(6x-5)

u = 6x-5 du/dx=6

1/6 int (du/dx)/u dx

=1/6 ln | u | + c

= 1/6 ln |6x-5| + c

confidence rating #$&*: 3

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Given Solution:

`a du/dx is the derivative of 6x-5, so du/dx = 6

If we let u = 6x - 5 then du = 6 dx so dx = 1/6 du and the integral becomes that of 1/u * du/6 = 1/6 (1/u du)

The integral of 1/6 * 1/u du is 1/6 * ln|u| + c

Substituting u = 6x - 5 we get the final result

int(1 / (6x - 5) = 1/6 * ln|6x-5|

**

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Self-critique (if necessary): ok

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Self-critique Rating: 3

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Question: `q 5.3.22 (was 5.3.5) (previously problem 5.3.22 (was 5.3.20)) integral of x/(x^2+4) by Log Rule

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Your solution:

x/(x^2+4)

int x/(x^2+4)

u = x^2 + 4

du/dx = 2x

˝ int ((du/dx)/ u) dx

= ˝ (ln |u| + c)

=1/2 ln |x^2 + 4| + c

confidence rating #$&*: 3

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Given Solution:

`a If we let u = x^2 + 4 we get du/dx = 2x so that the x in the numerator is 1/2 du/dx.

The integral of x / (x^2 + 4) is the integral of 1/2 * ( 2x / (x^2+4) ) = 1/2 (1/u du/dx).

The general antiderivative is therefore 1/2 ln(u) + c = 1/2 ln |x^2+4| + c. **

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `qWhat is the derivative of your result?

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Your solution:

Derivative of 1/2 ln |x^2 + 4| + c is x/(x^2+4)

ln(x^2+4) * (1/2)

=1/2 * 2x * 1 / (x^2 + 4)

=x / (x^2 + 4)

confidence rating #$&*:

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Given Solution:

`a The derivative of ln(x^2+4) * (1/2) is 1/2 * 2x * 1 / (x^2 + 4) or x / (x^2 + 4). This confirms that ln(x^2+4) * (1/2) is a solution to the equation.

The general antiderivative is of course ln(x^2+4) * (1/2) + c. **

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Self-critique (if necessary):

ok

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Self-critique Rating: e

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Question: `q 5.3.28 (was 5.3.7) (previously 5.3.28 (was 5.3.24) (was 5.3.24) ) integral of e^x/(1+e^x) by Log Rule

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Your solution:

e^x/(1+e^x)

u = 1+ e^x

du/dx = e^x

int ((du/dx) / u) dx

=ln | u | + c

=ln |1+ e^x) + c

confidence rating #$&*: 3

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Given Solution:

`a let u = 1 + e^x. Then du/dx = e^x.

We are therefore integrating 1 / (1 + e^x) * e^x, which is 1/u du/dx.

The antiderivative is ln |u| + c = ln | 1 + e^x | + c. **

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `q 5.3.9 (previously 5.3.46 (was 5.3.34) (was 5.3.34) ) integral of (6x + e^x) `sqrt( 3x^2 + e^x)

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Your solution:

6x + e^x) `sqrt( 3x^2 + e^x)

u = 3x^2 + e^x

du/dx = 6x + e^x

du = 6x + e^x dx

int (u)^1/2 * du

(u^3/2)/(3/2) + c

=2/3u^3/2 + c

=2/3 (3x^2+e^x)^3/2 + C

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a Here are two detailed solutions:

(6x + e^x) `sqrt( 3x^2 + e^x) = `sqrt(u) * du/dx = u^(1/2) du/dx.

The antiderivative is thus

2/3 u^(3/2) = 2/3 (3x^2 + e^x)^(3/2).

Alternatively

If u = 3x^2 + e^x then du = 6x + e^x and we have the integral of `sqrt(u) du, which is just

2/3 u^(3/2) + c = 2/3 (3x^2 + e^x)^(3/2) + c. **

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Self-critique (if necessary):

ok

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Self-critique Rating: 3

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Question: `q 5.3.58 (previously 5.3.11) (previously 5.3.58 (was 5.3.54) (was 5.3.52) ) dP/dt = -125 e^(-t/20), t=0, P=2500 and interpretation.

Give your complete solution.

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Your solution:

dP/dt = -125 e^(-t/20)

dp = -125 e^(-t/20) dt

int dp = int (-125 e^(-t/20) dt) u = -t/20 du = -20dt

p = 2500 e^(-t/20) + c

2500 = 2500 e^(0/20) + c

0 = c

solutions = p = 2500 e^(-t/20)

b. population after 15 days:

p = 2500e^(-15/20)

p = 1180.92

1180 trout

c. 1/2 = 2500 e^(-t/20)

.5/2500 = e^(-t/20)

ln (.5/2500) = -t/20

20 ln (.5/2500) = -t

-170.34 = -t

170 days = t

confidence rating #$&*: 3

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Given Solution:

`a If dP/dt = -125 e^(-t/20) then dp = -125 e^(-t/20) dt. Integrating both sides we get

p = 2500 e^(-t/20) + c ( to integrate the right-hand side start with u = -t / 20, etc.

If p = 2500 when t = 0 we have

2500 = 2500 e^(-0/20) + c so

2500 = 2500 + c and c = 0.

The final solution is thus

p = 2500 e^(-t/20)

After 15 days the population is p(15) = 2500 e^(-15/20) = 1000, give or take a couple hundred (you can evaluate the expression).

All the trout are considered dead when the population is below 1/2. So you need to solve 1/2 = 2500 e^(-t/20) for t.

Dividing both sides of this equation by 2500 then taking the natural log of both sides you get

-t/20 = ln( 1/2500 ) so

t = -20 * ln (1/2500) = -11 or -12 or so.

Thus t is about 200 days, give or take a little.

Alternative reasoning of the particular solution:

If u = -t/20 then e^u du/dt = e^(-t/20) * -1/20. -125 e^(-t/20) is 2500 * ( -1/20 e^(-t/20) ) = 2500 e^u du/dx.

The integral is 2500 e^u + c = 2500 e^(-t/20) + c.

If t = 0, P=2500 then 2500 = 2500 e^0 + c = 2500 + c, so c = 0. Thus the particular solution is

P = 2500 e^(-t/20).

Alternative solution for the time when all trout are dead:

2500 e^(-t/20) < .5 means

e^(-t/20) < .0002 so -t/20 < ln(.0002) so

-t < ln(.0002) * 20 so

-t < -170.34 and

t > 170.34.

The probability is that all trout are dead by day 171.

STUDENT QUESTION: I couldn't figure out the time for all the trout to die because the ln 0 is undefined

** When the population falls below 1/2 of a fish it rounds off to 0 and you assume that all the trout are dead.

You can think of this in terms of probability. The function doesn't really tell us the precise number but the probable number. When the probability is againt that last fish being alive we figure that it's most likely dead. **

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Self-critique (if necessary):

I had similar question as previous student, I thought p should = 0 when all trout were dead but reading through reasoning ˝ makes sense and I was comfortable solving problem when p = 1/2

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Self-critique Rating: 3

@&

Good.

The problem with p = 0 is that the value of p would approach 0 as an asymptote, but never actually reach 0.

*@

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#