If your solution to a stated problem does not match the given solution, you should self-critique per instructions at  

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution: If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

014.

Note the document qa_ac2_14.htm , which gives an introduction to the logic and mechanics of integration by substitution.

Question: `qQuery problem 6.1.5 (was 6.1.4) integral of (2t-1)/(t^2-t+2)

Your solution:

Confidence Assessment:

Given Solution: `a To integrate (2t-1) / (t^2-t+2) use u = t^2 - t + 2. We find that du = (2 t - 1) dt, which is there just waiting for you in the integrand.

This gives you integrand du / u.

The integral is ln | u | + c. Substituting we get

int( (2t-1) / (t^2 - 1 + 2) with respect to t) = ln | t^2 - t + 2 | + c.

The absolute value is important because t^2 - t + 2 can be negative.

STUDENT QUESTION:

Having a hard time still seeing how to get started

INSTRUCTOR RESPONSE

Understood.  This is a common problem. 

Once you see what to do it's fairly straightforward, but how do you see what needs to be done?

Generally I recommend checking to see if any other part of the integrand has a derivative which is equal to, or a multiple of, a factor of the integrand. 

Just what does this mean?

In the present case the integrand is (2t-1)/(t^2-t+2). 

What are the factors of the integrand?

The integrand has only one factor, which is (2 t - 1).

What are the 'other parts' of the integrand, and what are their derivatives?

The only thing you have besides the factor (2 t - 1) is the denominator t^2 - t + 2.

 

The derivative of this 'part' is 2 t - 1.

Do any of those derivatives match any of the factors?

Yes.  The derivative is 2 t - 1, and the only factor is 2 t - 1.

If the answer to the above is 'yes', then let u be the part whose derivative is equal to the factor and proceed.

Ok, so u = t^2 - t + 1.

 

Thus du/dt = 2 t - 1, so du = (2 t - 1) dt.

 

The expression (2 t - 1) / (t^2 - t + 2) dt has 'numerator' (2 t - 1) dt and denominator u.  So the expression can be written du / u.

 

An antiderivative of du / u is ln | u |, and the general antiderivative is ln | u | + c.

 

Thus we get our general solution ln | t^2 - t + 2 | + c.

Self-critique (if necessary):

Self-critique Rating:

Question: `qQuery problem 6.1.32 (was 6.1.26) integral of 1 / (`sqrt(x) + 1)

Your solution:

Confidence Assessment:

Given Solution: `a If we let u = (sqrt x) + 1 we can solve for x to get

x = (u-1)^2 so that

dx = 2(u-1) du.

So the integral of 1 / (`sqrt(x) + 1) dx becomes the integral of (2 ( u - 1 ) / u) du.

Integrating ( u - 1) / u with respect to u we express this as

( u - 1) / u = (u / u) - (1 / u) = 1 - (1 / u). An antiderivative is u - ln | u |.

Substituting u = (sqrt x) + 1 and adding the integration constant c we end up with

(sqrt x) + 1 - ln | (sqrt x) + 1 | + c.

Our integral is of 2 (u-1)/u, double the expression we just integrated, so our result will also be double. We get

2 (sqrt x) + 2 - 2 ln | (sqrt x) + 1 | + c.

Since c is an arbitrary constant, 2 + c is also an arbitrary constant so the final solution can be expressed as

2 (sqrt x) - 2 ln | (sqrt x) + 1 | + c.

Self-critique (if necessary):

Self-critique Rating:

Question: `qquery problem 6.1.56 (was 6.1.46) area bounded by x (1-x)^(1/3) and y = 0

Your solution:

Confidence Assessment:

Given Solution: `a Begin by sketching a graph to see that there is a finite region bounded by the graph and by y = 0. Then find the points where the graph crosses the line y = 0 but finding where the expression takes the value 0:

x(1-x)^(1/3) = 0 when x = 0 or when 1-x = 0. Thus the graph intersects the x axis at x = 0 and x = 1.

We therefore integrate x (1-x)^(1/3) from 0 to 1.

We let u = 1-x so du = dx, and x = 1 - u.

This transforms the integrand to (1 - u) * u^(1/3) = u^(1/3) - u^(4/3), which can be integrated term by term, with each term being a power function.

Our antiderivative is 3/4 u^(4/3) - 3/7 u^(7/3), which translates to 3/4 (1-x)^(4/3) - 3/7 ( 1-x)^(7/3).

The result is 9/28 = .321 approx..

Self-critique (if necessary):

Self-critique Rating:

Question: `qQuery problem P = int(1155/32 x^3(1-x)^(3/2), x, a, b).

Your solution:

Confidence Assessment:

Given Solution: `a For reference

int(1155/32 x^3(1-x)^(3/2), x, a, b) means 'the integral of 1155 / 32 x^3 ( 1 - x)^(3/2), integrated with respect to x

between limits x = a and x = b'.

That would be written with an integral sign with limits a and b, then 1155/32 x^3(1-x)^(3/2) dx.

It's easiest to integrate if you change the variable to u = 1 - x, which changes the integrand to 1155/32 (1 + u)^3 *

u^(3/2). Expand the cube and multiply through by u^(3/2) to get a sum of four power functions, easily integrated

The result is of course a bit messy:

First we expand the cube:  (1 + u)^3 = 1 + 3 u + 3 u^2 + u^3, so

1155/32 ((1 + u)^3)(u^(3/2)) =

1155 / 32 ( 1 + 3 u + 3 u^2 + u^3) * u^(3/2) =

1155 / 32 ( u^(3/2) + 3 u^(5/2) + 3 u^(7/2) + u^(9/2)).

An antiderivative is

1155 / 32 ( 2/5 u^(5/2) + 6/7 u^(7/2) + 6/9 u^(9/2) + 2/11 u^(11/2) ).

Since u = 1 - x the limits on the u integral would be 1 - a and 1 - b.

The result would therefore be

1155 / 32 ( 2/5 (1-b)^(5/2) + 6/7 (1-b)^(7/2) + 6/9 (1-b)^(9/2) + 2/11 (1-b)^(11/2) ) - 1155 / 32 ( 2/5 (1-a)^(5/2) + 6/7 (1-a)^(7/2) + 6/9 (1-a)^(9/2) + 2/11 (1-a)^(11/2) ), which is slightly simplified to

1155 / 32 ( 2/5 (1-b)^(5/2) + 6/7 (1-b)^(7/2) + 6/9 (1-b)^(9/2) + 2/11 (1-b)^(11/2) ) - 2/5 (1-a)^(5/2) - 6/7 (1-a)^(7/2) - 6/9 (1-a)^(9/2) - 2/11 (1-a)^(11/2) ).

Factoring (1 - b)^(5/2) from the first four terms and (1 - a)^(5/2) from the last four terms, and simplifying yields

((1 - a)^(5/2)(105 a^3 + 70 a^2 + 40 a + 16) - (1 - b)^(5/2) (105 b^3 + 70 b^2 + 40 b + 16))/16.

STUDENT COMMENT

I understand the integration but the expanding of the cube is were I got lost.

INSTRUCTOR RESPONSE

Use the distributive and commutative laws to expand the cube:

(1 + u)^3 = (1 + u) * (1 + u) ^2

(1 + u)^2

= (1 + u) ( 1 + u)

= 1 ( 1 + u) + u * (1 + u)

= 1 + u + u + u^2

= u^2 + 2 u + 1, so

(1 + u)^3

= (1 + u) * (1 + u) ^2

= (1 + u) * (u^2 + 2 u + 1)

= 1 * (u^2 + 2 u + 1) + u * (u^2 + 2 u + 1)

= u^2 + 2 u + 1 + u^3 + 2 u^2 + u

= u^3 + 3 u^2 + 3 u + 1.

Self-critique (if necessary):

Self-critique Rating:

Question: `qWhat is the probability that a sample will contain between 0% and 25% iron?

Your solution:

Confidence Assessment:

Given Solution: `a The probability of an occurrence between 0 and .25 is found by integrating the expression from x = 0 to x = .25:

Let u = 1-x so du = -dx and x = 1-u.

Express in terms of u:

-(1155/32) * int ( (1-u)^3 (u)^(3/2) du )

Expand the integrand:

-(1155/32) * int( (1 - 3 u + 3 u^2 - u^3) * u^(3/2) ) =

-1155/32 * int( u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) ) .

An antiderivative of u^(3/2) - 3 u^(5/2) + 3 u^(7/2) - u^(9/2) is 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) . So we obtain for the indefinite integral

-1155/32 * ( 2/5 u^(5/2) - 3 * 2/7 u^(7/2) + 3 * 2/9 * u^(9/2) - 2/11 u^(11/2) ).

Express in terms of x:

-1155/32 * ( 2/5 ( 1 - x )^(5/2) - 3 * 2/7 ( 1 - x )^(7/2) + 3 * 2/9 * ( 1 - x )^(9/2) - 2/11 ( 1 - x )^(11/2) )

Evaluate this antiderivative at the limits of integration 0 and .25 .

You get a probability of .0252, approx, which is about 2.52%, for the integral from 0 to .25; this is the probability of a result between 0 and 25%.

To get the probability of a result between 50% and 100% integrate between .50 and 1. You get .736, which is 73.6%

Self-critique (if necessary):

Self-critique Rating:

`qQuery Add comments on any surprises or insights you experienced as a result of this assignment.