Assignment 19

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course MTH 272

12/10 about 4:45 pm

019.

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Question: `qQuery problem 6.4.16 use table to integrate x^2 ( ln(x^3) )^2

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Your solution:

U=x^3 du=3x^2 x^2/(3x^2) = 1/3

Int. (x^2(ln(x^3))^2) = 1/3*[u*(2 - 2*ln u + (ln u)^2)]+c

=1/3*x^3*(2 - 2*ln x^3 + (ln x^3)^2) + c

=((x^3*(2 - 2*ln x^3 + (lnx^3)^2))/3) + c

confidence rating #$&*:3

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Given Solution:

`a Let u = x^3 so du = 3x^2 dx. The x^2 dx in the integral is just 1/3 du.

You therefore have the integral of 1/3 ( ln(u) )^2 du.

The table should have something for ( ln(u) ) ^ n.

In any case the integral of ln(u)^2 with respect to u is u ln(u)^2 - 2 u ln(u) + 2 u.

With the substitution u = x^3 you would be integrating (ln u)^2 * du/3, which would give you

u [ 2 - 2 ln u + (ln u)^2 ] / 3, which translates to

x^3 ( 2 - 2 ln(x^3) + (ln(x^3) ) ^ 2 ) / 3.

DER: int( (ln(u)^2) = u•LN(u)^2 - 2•u•LN(u) + 2•u. Then for increasing powers of n int( ln(u)^n) gives us:

u•LN(u)^3 - 3•u•LN(u)^2 + 6•u•LN(u) - 6•u then

u•LN(u)^4 - 4•u•LN(u)^3 + 12•u•LN(u)^2 - 24•u•LN(u) + 24•u and

u•LN(u)^5 - 5•u•LN(u)^4 + 20•u•LN(u)^3 - 60•u•LN(u)^2 + 120•u•LN(u) - 120•

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `qQuery problem 6.3.52 (7th edition 6.4.46) use table to integrate x ^ 4 ln(x) then check by integration by parts

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Your solution:

Table

U=x n=4

Int.(x^4 * ln(x) dx) = int. (u^n*ln u du) = (u^(n+1)/(n+1)^2)*[-1 + (n+1)*ln u] + c

=(x^5/25)*[-1 + 5*ln x] + c = (5x^5*lnx - x^5)/25 + c or (x^5*ln x)/5 - (x^5/25) + c

Integration by parts

U=ln x du= 1/x dx dv=x^n dx v=(x^(n+1)/(n+1)

Ln x *(x^(n+1)/(n+1)) - int. (x^n+1/(n+1) * 1/x dx

Ln x *(x^(n+1)/(n+1)) - int. (x^n dx/(n+1))

Ln x *(x^(n+1)/(n+1)) - (x^(n+1)/(n+1)^2)

(x^(n+1)/(n+1))*(ln x - 1/(n+1))

(x^5/5)*(ln x - 1/5)=((x^5 * ln x) / 5) - x^5/25 + c

confidence rating #$&*:3

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Given Solution:

`a Integration by parts on x^n ln(x) works with the substitution

u = ln(x) and dv = x^n dx, so that

du/dx = 1/x and v = x^(n+1) / n, giving us

du = dx / x and v = x^(n+1) / (n + 1).

Thus our integral is

u v - int( v du) =

ln(x) * x^(n+1) / (n + 1) - int ( x^(n+1)/(n+1) * dx / x) =

ln(x) * x^(n+1)/( n + 1) - int(x^n dx) / (n+1) =

ln(x) * x^(n+1) / (n + 1) - x^(n+1) / (n+1)^2 =

x^(n+1) / (n+1) ( ln(x) - 1(n+1)).

This should be equivalent to the formula given in the text.

For n = 4 we get

x^(4 + 1) / (4 + 1) ( ln(x) - 1 / (4 + 1)) =

x^5 / 5 (ln(x) - 1/5). *&*&

(x^5/25)(4 ln x) + C

Using integration by parts:

(x^5/5) ln x - (x^5/25)

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Self-critique (if necessary):OK

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Self-critique Rating:OK

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Question: `qQuery problem 6.4.63 profit function P = `sqrt( 375.6 t^2 - 715.86) on [8,16].

( 8th edition problem is 6.2.61 and the function is sqrt(.000645 t^2 + .1673) on (2, 5) )

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Your solution:

u^2=375.6t^2 u=19.38t a^2=715.86 a=26.76

(1/2)*(19.38*sqrt(375.6t^2 - 715.86) - 715.86*ln|19.38t + sqrt(375.6t^2-715.86|)

Plugging in the t values and subtracting I get 45595.7-9787.94=35,807.7

confidence rating #$&*:3

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Given Solution:

`a To get the average net profit integrate the profit function over the given interval and divide by the length of the interval.

The integrand is sqrt(375.6 t^2 - 715.86), which is of the form `sqrt( u^2 +- a^2).

• By the table the integral is

1/2 u sqrt(u^2 +- a^2) -+ a^2 ln(u + sqrt(u^2 +- a^2)) + c.

u^2 = 375.67 t^2 so u = `sqrt(375.67) * t = 19.382 t approx.

Similarly a = `sqrt(715.86) = 26.755 approx..

Substituting our values of a and u we find that integral(sqrt(375.6 t^2 - 715.86), t from 8 to 16) will be about 1850. Dividing this by the length 8 of the interval gives us the average value, which is about 1850 / 8 = 230.

** THE FUNCTION IS CLOSE TO THE LINEAR FUNCTION 19.4 t. The 715.86 doesn't have much effect when t is 8 or greater so the function is fairly close to P = 19 t. This approximation is linear so its average value will occur at the midpoint t = 12 of the interval. At t = 12 we have P = 19 * 12 = 230, approx.

8TH EDITION INTEGRAL

The function given in the 8th edition leads to the integral

int( sqrt(.000645 t^2 + .1673) dt, t from 2 to 5), which is of the form `sqrt( u^2 +- a^2).

u^2 = .000645 t^2 so u = `sqrt(.000645) * t = .08 t approx.

Similarly a = `sqrt(.1673) = .41 approx..

• By the table the integral is

1/2 u sqrt(u^2 +- a^2) -+ a^2 ln(u + sqrt(u^2 +- a^2)) + c.

u^2 = 375.67 t^2 so u = `sqrt(375.67) * t = 19.382 t approx.

Similarly a = `sqrt(715.86) = 26.755 approx..

Substituting our values of a and u we find that

integral(sqrt(.000645 t^2 + .1673), t from 2 to 5) will be about 1850. Dividing this by the length 8 of the interval gives us the average value, which is about 1850 / 8 = 230.

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Self-critique (if necessary):I got a different answer than you. I thought at first, I might have needed a plus in front of 715.86*ln| | but when I solved that way I still did not get the right answer. If I do need a plus there, how do you know which to follow off the table + or -?

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Self-critique Rating:3

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The value of the integral, correct to the nearest whole number, is 1848.

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Preceding the a^2 is +-.

Any term in the formula preceded by +- will have the same sign as a^2.

Any term preceding by -+ will have the sign opposite that of a^2.

So your expression should have been

(1/2)*(19.38*sqrt(375.6t^2 + 715.86) - 715.86*ln|19.38t + sqrt(375.6t^2+715.86|)

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

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Self-critique (if necessary):

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Self-critique rating:

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

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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. &#