Assignment 4

course MTH 174

I am resubmitting this per the instructions in orientation about resubmitting work that has not been posted within a day.

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Physics II

09-22-2008

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19:40:17

query problem 7.2.12 (3d edition #11) antiderivative of sin^2 x

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RESPONSE -->

This one was in section 7.3 and was #12. Section 7.2 was Integration by substitution and this seems better fitted to integration by parts.

Here is the solution that I have:

Using integration by parts (uv - integral uv' ):

integral sin^2 x

integral sin x * sin x

let u = sin x and v' = sin x

du/dx = cos x so du = cos x dx

v = -cos x (+ C omitted here)

uv - integral uv'

sin x * - cos x - integral - cos^2 x dx

-sin x cos x + integral (1 - sin^2 x) dx

-sin x cos x + integral 1 dx - integral sin^2 x

We see sin^2 x again. This integral is equation above is supposed to be equal to integral sin^2 x

-sin x cos x + integral 1 dx - integral sin^2 x = integral sin^2 x

-sin x cos x + x + C = 2 integral sin^2 x

-1/2 sin x cos x + x/2 + C = integral sin^2 x

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19:40:33

what is the requested antiderivative?

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RESPONSE -->

-1/2 sin x cos x + x/2 + C

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19:42:29

What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?

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RESPONSE -->

The sin^2 x was broken into sin x and sin x. The technique used was integration by parts. The trick here was using the fact that integral of sin^2 x appeared again in the problem allowing the movement of it to the other side of the equation.

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19:58:35

query problem 7.2.16 antiderivative of (t+2) `sqrt(2+3t)

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RESPONSE -->

This one also appeared in section 7.3 in the integration by parts.

I used u = t+2 and v' = (2 + 3t)^(1/2)

du/dt = 1

substituting in w = 2+3t and dw/dt = 3, so 1/3dw = dt

v' = 1/3 w^(1/2) dw

v = 1/3 * 2/3 w^(3/2)

v = 2/9 (2 + 3t)^(3/2)

uv - integral uv'

(t+2)[2/9 (2 + 3t)^(3/2)] - integral 1 * 2/9 (2 + 3t)^(3/2) dt

Again, letting w = 2+3t and 1/3dw = dt

2/9(t +2)(2 + 3t)^(3/2) - 2/9 integral 1/3 w^(3/2) dw

2/9(t +2)(2 + 3t)^(3/2) - 2/27 * 2/5 (2 + 3t)^(5/2) + C

2/9(t +2)(2 + 3t)^(3/2) - 4/135(2 + 3t)^(5/2) + C

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19:58:46

what is the requested antiderivative?

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RESPONSE -->

2/9(t +2)(2 + 3t)^(3/2) - 4/135(2 + 3t)^(5/2) + C

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20:00:05

What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?

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RESPONSE -->

I broke it into u = t+2 and v' = (2 + 3t)^(1/2) and used integration by parts. Substitution was also used in finding the antiderivative for v and in the integral portion of the integration by parts.

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20:10:19

query problem 7.2.27 antiderivative of x^5 cos(x^3)

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RESPONSE -->

This one also was in 7.3. Again, I used integration by parts (uv - integral uv').

integral x^5 cos(x^3) dx

let u = x^3

du/dx = 3x^2

v' = x^2 cos(x^3)

Substituting w = x^3, dw/dx = 3x^2 so 1/3dw = x^2dx

v' = 1/3 cos w dw

v = 1/3 sin (x^3)

uv - integral uv'

x^3 * 1/3 sin (x^3) - integral 3x^2 * 1/3 sin (x^3) dx

x^3 * 1/3 sin (x^3) - integral x^2 sin (x^3) dx

Again, substituting w = x^3:

dw/dx = 3x^2 so 1/3dw = x^2dx

x^3 * 1/3 sin (x^3) - integral 1/3 sin (w) dw

1/3 x^3 sin (x^3) + 1/3 cos (x^3) + C

Note that you could also start with a simple substitution u = x^3. This gives you du = 3 x^2 dx; factoring this out of x^5 cos(x^3) dx you get (x^3 cos(x^3) ) (x^2 dx), which in terms of u is (u cos(u)) * du/3. This, of course, would be integrated by parts so this arrangement is probably no simpler than your direct approach, but it does provide an interesting alternative.

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20:10:28

what is the requested antiderivative?

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RESPONSE -->

1/3 x^3 sin (x^3) + 1/3 cos (x^3) + C

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20:12:39

What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?

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RESPONSE -->

let u = x^3 and v' = x^2 cos(x^3) The key was recognizing that the derivative of x^3 is 3x^2 after that, it was straight forward. The substitution of w = x^3 was used in the finding of the antiderivative of v' and in the integral in the integration by parts.

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20:49:51

query problem 7.2.52 (3d edition #50) f(0)=6, f(1) = 5, f'(1) = 2; find int( x f'', x, 0, 1).

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I am having problems visualizing this one. I'll describe what I have sketched out on my paper I have a trapezoidal graph of x vs t with a point ot (0,6) and a point at (1,5) and a slope between the two. At point (1,5) the slope is 2.

You can find this one using integration by parts:

Let u=x and v' = f''(x). Then

u'=1 and v=f'(x).

uv-integral of u'v is thus

xf'(x)-integral of f'(x)

Integral of f'(x) is f(x). So antiderivative is

x f ' (x)-f(x), which we evaluate between 0 and 1. Using the given values we get

1 * f'(1)- (f(1) - f(0)) =

f ‘ (1) + f(0) – f(1) =

2 + 6 - 5 = 3.

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20:50:12

What is the value of the requested integral?

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I am not sure.

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20:51:04

How did you use integration by parts to obtain this result? Be specific.

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I was unable to find the result.

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20:51:11

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

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20:51:16

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

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