Assignment 4

course Mth 174

«¸ä£¦]°“„âñœ˜–ë¾|jô­øassignment #004

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Physics II

01-30-2007

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17:22:18

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

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17:35:04

what is the requested antiderivative?

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RESPONSE -->

Antiderivative of sin^2 theta d theta:

u = sin theta u' = cos theta

v' = sin theta v = -cos theta

Integral of sin^2 theta = -sin theta cos theta - (integral of -cos^2 theta d theta) + C

Substitute (1 - sin^2 theta) for cos^2 theta:

Integral of sin^2 theta = -sin theta cos theta + integral of (1 - sin^2 theta d theta) + C

Integral of sin^2 theta = -sin theta cos theta + integral of (1 d theta) - integral of (sin^2 theta d theta) + C

Adding integral of (sin^2 theta d theta) to both sides:

2 * integral of sin^2 theta d theta = -sin theta cos theta + theta + C

Dividing both sides by 2:

Integral of sin^2 theta d theta = (-sin theta cos theta/2) + theta/2 + C

Good. Note the two alternative solutions given below, one of which follows the same scheme as yours:

We have a choice here of integration by parts or 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

Integration by parts:

Let u= sinx and dv = sinx dx. Then v = -cos(x) and

u v – int(v du) = -sinx cosx + int (cos^2 x)

= -1/2 sinx cosx + 1/2 x + C

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17:35:19

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

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RESPONSE -->

See previous response

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17:36:47

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

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18:06:00

what is the requested antiderivative?

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RESPONSE -->

Antiderivative of (t+2)*[sqrt(2+3t)] dt

Rewrite: (t+2)*(2+3t)^(1/2)

u = t+2 u' = 1

v' = (2+3t)^(1/2) v = (2/9)*(2+3t)^(3/2)

= (t+2)*(2/9)*(2+3t)^(3/2) - integral of [(2/9)(2+3t)^(3/2) dt] + C

= (2/9)*(t+2)*(2+3t)^(3/2) -(2/9)* integral of [(2+3t)^(3/2) dt] + C

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

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

Good solution. For reference and comparison:

If you use

u=t+2

u'=1

v'=(2+3t)^(1/2)

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

then you get

2/9 (t+2) (3t+2)^(3/2) - integral( 2/9 (3t+2)^(3/2) dt ) or

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

2/9 (t+2) (3t+2)^(3/2) - 4/135 (3t+2)^(5/2). Factoring out (3t + 2)^(3/2) you get

(3t+2)^(3/2) [ 2/9 (t+2) - 4/135 (3t+2) ] or

(3t+2)^(3/2) [ 30/135 (t+2) - 4/135 (3t+2) ] or

(3t+2)^(3/2) [ 30 (t+2) - 4(3t+2) ] / 135 which simplifies to

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

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18:06:11

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

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RESPONSE -->

See previous response

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19:04:31

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

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19:13:28

what is the requested antiderivative?

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RESPONSE -->

Antiderivative of x^5*cos(x^3) dx

u = x^3 du = 3x^2 dx du/3 = x^2 dx

Substitute:

= integral of u * du/3 * cos u

= (1/3) integral of u cos u du

w = u w' = 1

v' = cos u v = sin u

Substitute using integration by parts:

= (1/3) [u sin u - integral of sin u du] + C

= (1/3) [u sin u - (-cos u)] + C

= (1/3) u sin u + (1/3) cos u + C

Substitute back:

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

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19:13:40

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

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RESPONSE -->

See previous response

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19:46:05

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

u = x u' = 1*dx = du

Since u = x, substitute u for x

v' = f''(u) v = f'(u)

Integration by parts:

= [u*f'(u)] (evaluated from 0 to 1) - integral of f'(u) du (evaluated from 0 to 1)

= [u*f'(u) - f(u)] evaluated from 0 to 1

Substitute back:

= [x*f(x) - f(x)] evaluated from 0 to 1

Evaluate:

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

= (2 - 5) - (0 - 6)'

= 3

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19:46:16

What is the value of the requested integral?

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See previous response

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19:46:31

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

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See previous response

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19:51:16

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

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I found that integration by parts sometimes requires several tries to find a substitution that work, although as I got more practice, certain patterns emerged that helped in choosing the proper substitutions. Checking is essential for catching errors!

Absolutely so. Excellent insight, as usual.

Very good work.