Assignment 4

course MTH 174

FPӡ|֟assignment #004

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Ճ߉mְWb

Physics II

11-05-2007

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14:32:17

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

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16:22:37

what is the requested antiderivative?

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RESPONSE -->

(x / 2) - (sin(x)*cos(x)) / 2

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16:26:06

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

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RESPONSE -->

I used the formula:

antiderivative of sin^n(x) dx =

(-1 / n) * sin^(n-1)(x) * cos(x) + ((n -1) / n) * antiderivative of sin^(n-2)(x) dx, n positive

The formula from the table is not a legitimate means to use in Section 7.2. You need to use substitution to obtain the formula yourself.

In general unless a problem states that tables are permissible, they are not.

Two possible solutions are shown below:

Good student solution:

The answer is -1/2 (sinx * cosx) + x/2 + C

I arrived at this using integration by parts:

u= sinx u' = cosx

v'= sinx v = -cosx

int(sin^2x)= sinx(-cosx) - int(cosx(-cosx))

int(sin^2x)= -sinx(cosx) +int(cos^2(x))

cos^2(x) = 1-sin^2(x) therefore

int(sin^2x)= -sinx(cosx) + int(1-sin^2(x))

int(sin^2x)= -sinx(cosx) + int(1) int(sin^2(x))

2int(sin^2x)= -sinx(cosx) + int(1dx)

2int(sin^2x)= -sinx(cosx) + x

int(sin^2x)= -1/2 sinx(-cosx) + x/2

INSTRUCTOR COMMENT: This is the appropriate method to use in this section.

You could alternatively use trigonometric identities such as

sin^2(x) = (1 - cos(2x) ) / 2 and sin(2x) = 2 sin x cos x.

Solution by trigonometric identities:

sin^2(x) = (1 - cos(2x) ) / 2 so the antiderivative is

1/2 ( x - sin(2x) / 2 ) + c =

1/2 ( x - sin x cos x) + c.

note that sin(2x) = 2 sin x cos x.

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19:30:32

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

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RESPONSE -->

antiderivative of (t+2) `sqrt(2+3t)

u = (t+2) v' = `sqrt(2+3t)

antiderivative of uv' dt = uv - antiderivative of u'v dt

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19:32:04

what is the requested antiderivative?

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RESPONSE -->

2(3x+2)^(3/2) * (9x+26) / 135

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19:50:43

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

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RESPONSE -->

antiderivative of (t+2) `sqrt(2+3t)

u = (t+2) v' = `sqrt(2+3t)

antiderivative of uv' dt = uv - antiderivative of u'v dt

antiderivative of (t+2) `sqrt(2+3t) =

(t+2)2(2+3t)^(3/2) / 9 - antiderivative of 2(2+3t)^(3/2) / 9

antiderivative of 2(2+3t)^(3/2) / 9 = (2/9) antiderivative of (2+3t)^(3/2)

antiderivative of (2+3t)^(3/2) = 2(2+3t)^(5/2) / 15

(t+2)2(2+3t)^(3/2) / 9 - (2/9) (2(2+3t)^(5/2) / 15) + C =

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

2(3t+2)^3/2 * (9t+26) / 135

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23:13:56

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

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23:16:37

what is the requested antiderivative?

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RESPONSE -->

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

I don't understand how to do this one. Please explain a viable method.

let u = x^3, v' = x^2 cos(x^3).

Then u' = 3 x^2 and v = 1/3 sin(x^3) so you have

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

Now let u = x^3 so du/dx = 3x^2. You get

1/3 * x^3 sin(x^3) - 1/3 * int( sin u du ) = 1/3 (x^3 sin(x^3) + cos u ) = 1/3 ( x^3 sin(x^3) + cos(x^3) ).

It's pretty neat the way this one works out, but you have to try a lot of u and v combinations before you come across the right one.

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23:16:40

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

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RESPONSE -->

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Oy҅IJҙs몷

assignment #004

Ճ߉mְWb

Physics II

11-09-2007

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20:40:15

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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RESPONSE -->

f(0) = 6, f(1) = 5

f'(1) = 2

Integral from 0 to 1 of x * f''(x) dx

u = x, v' = f''(x)

u' = 1, v = f'(x)

x * f'(x) - integral from 0 to 1 of f'(x) dx =

x * f'(x) - f(x), evaluated from 0 to 1 =

(f'(1) - f(1)) - (0 - f(0)) = (2 - 5) - (0 - 6) = 3

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20:40:21

What is the value of the requested integral?

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RESPONSE -->

3

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21:02:36

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

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RESPONSE -->

f(0) = 6, f(1) = 5

f'(1) = 2

Integral from 0 to 1 of x * f''(x) dx

u = x, v' = f''(x)

u' = 1, v = f'(x)

x * f'(x) - integral from 0 to 1 of f'(x) dx =

x * f'(x) - f(x), evaluated from 0 to 1 =

(f'(1) - f(1)) - (0 - f(0)) = (2 - 5) - (0 - 6) = 3

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21:02:40

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

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21:02:45

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

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Good work.

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Let me know if you have questions. &#