Query 7

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course Mth 272

007. `query 7If 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: `q5.2.3 (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:

(1+ 1/3 x^2)

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You don't indicate what you were thinking or how you got this.

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1+ 1/9 x + C or ((1-x^3)^3)/3+C

confidence rating #$&*: 2

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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 don’t understand where the term -1/9 comes into play in this equation and why there is not denominator for the u term.

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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: 2x^2+C

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The derivative of your result should equal the original function you integrated.

You don't indicate what you took the derivative of to get this, or how you followed the rules of derivatives to arrive at your answer.

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confidence rating #$&*: 1

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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): I’m not sure if you have to solve the problems for an answer to the variable given or just simplify/ rearrange the problem to general derivative form.

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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:

-40/(.02p-1)^2

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Again you need to indicate the details of your thinking and the steps you used to get this.

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confidence rating #$&*: 1

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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): Again, I am unsure whether to solve this equation or simplify the expression putting it in general form.

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Self-critique Rating: 3

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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:

-4e(.25x)+C

confidence rating #$&*: 2

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G=[[ven Solution:

`l]]le 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): I’m not quite sure why (.25x) remains negative, but this question was fairly basic.

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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:

(x-4)/5e^(x^2-8x)+C

confidence rating #$&*: 1

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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): Not necessary.

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Self-critique Rating: 2

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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: ln(6x-5)+C

confidence rating #$&*: 2.5

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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): I arrived at my solution because I thought once I took the derivative of (1/x) then it would become the ln of the problem.

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If you take the derivative of your answer you get 6 / (6x - 5), not 1 / (6x - 5).

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Self-critique Rating:

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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(x^2-4)]+c

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You don't indicate how you got this.

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confidence rating #$&*: 2

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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): I’m not sure where the ln comes into play and why it doesn’t with the previous problem.

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There was an error in the solution of the previous problem. I've corrected it above, and in the Query.

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Self-critique Rating: 2

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Question: `qWhat is the derivative of your result?

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Your solution:

Ln(x^2+4)*(1/2)+C

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): Not necessary

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Self-critique Rating: 3

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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:

Ln(1*e^x)+C

confidence rating #$&*: 2

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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): Not necessary.

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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)(3/2x)

confidence rating #$&*: 1

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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):

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Self-critique Rating: 1

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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:

2500e^(0/20)+C

confidence rating #$&*: 1

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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 didn’t think I needed to simplify my answer that much as displayed in the given solution in order to put my answer in general form.

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Self-critique Rating: 3

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@&

I suggest you check my notes then rework these problems and show more of your thinking, and the steps you've taken. In some cases I can see what you might have been thinking from your answers, but even then I need to see the process to be sure.

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