course Mth 272

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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:

let u = x^3

?????????? I would try to use formula 42 which is ʃ (ln u)^2 du = u [2-2 ln u + (ln u)^2] + C

But do not understand what to do with x^2

confidence rating #$&*1

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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):

??????? Not sure how you got x^2 = 1/3 du

Let u = x^3 so du = 3x^2 dx.

If we divide both sides of the equation du = 3 x^2 dx, we get

du / 3 = (3 x^2 dx) / 3, which simplifies to

1/3 du = x^2 dx.

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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:

Using the table

x^4 ln(x) d x we will use formula number 41

u^n * ln(u) du = u^(n+1) / (n+1)^2 [-1+(n+1)]*ln u + c

we let u = x and n = 4

x^4 ln(x) du = x^(4+1) / (4+1)^2 [-1+(4+1)] * ln(x) + c

= x^5 / 5^2 *4 ln(x) + C

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

Using intergration by parts:

let u = ln x du = 1/x dx dv = x^4 dx v = integral dv = integral x^4 dx = x^5 / 5

ʃ u * dv = uv - ʃ v dv

ʃx^4 ln(x) dx = x^5 / 5 * ln (x) - ʃ x^5 / 5 * 1 / x dx

= x^5 / 5 ln(x) – 1 / 5 ʃ x^4 dx

= x^5 / 5 * ln(x) – x^(4 +1) / (4+1)

= x^ 5 / 5 * ln(x) – (1 / 5) (x^5 / 5)

= x^5 / 5 * ln(x) – x^5 / 25 + C

confidence rating #$&*2

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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):

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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:

Use formula 21 from the table intergrals

ʃ sqrt(u^2 + a^2) du = ½( u *sqrt(u^2 + a^2) + a ln | u + sqrt (u^2 + a^2|) + c

let u^2 = sgrt(.00645*t^2) = 0.08t let a^2 = sqrt(.1673) = 0.41

= ½ (0.08t *sgrt (0.08t^2 +- 0.41^2) +- 0.41 ln| 0.08t + (0.08t^2 +- 0.41^2|) + C

sqrt(u^2 +- a^2) must match the expression
`sqrt( .000645 t^2 + .1673).
To do so, the + sign must be chosen from the +-, u^2 must be equal to 375.6 t^2 and a^2 must be equal to 715.86.
Thus
u^2 = .000645 t^2 so that u = sqrt(.000645 t^2) = sqrt(.000645) * sqrt(t^2) = .08 t, approx.
and
a^2 = .1673 so a = .41, approx..
We plug a = .41 and u = .08 t into ½( u *sqrt(u^2 + a^2) + a ln | u + sqrt (u^2 + a^2|) to get the antiderivative.
We evaluate this at 2 and 5, and subtract the results.

confidence rating #$&*0

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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.

Substituting our values of a and u, and choosing the positive sign from the +-, we find that and antiderivative is

1/2 u sqrt(u^2 + a^2) = a^2 ln(u + sqrt(u^2 + a^2)) = 1/2 (.08 t) sqrt( .64 t^2 + .167 ) + .167 ln (.08 t + sqrt( .64 t^2 + .167).

so that the integral is

integral(sqrt(.000645 t^2 + .1673), t from 2 to 5) will be about 1.26.

Dividing this by the length 8 of the interval gives us the average value, which is about 1.26 / 8 = .16, approx..

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Self-critique (if necessary):

???????The math got me on this one

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

??????? Not sure how x^2 is equal to 1 / 3 du in problem one

&#Be sure to see my note(s), inserted at various places in this document, and let me know if you have questions. &#