Assignment 4

course Mth 174

zXΒCӮassignment #004

د`SB

Physics II

06-05-2009

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18:47:01

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

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RESPONSE -->

antiderivative of sin^2 x=

integral sin(theta)sin(theta) = -sin(theta)cos(theta) - integral - cos(theta)cos(theta)=

cos^2 = 1 + cosx so,

integral sin(theta)sin(theta)= -sin(theta)cos(theta) + integral 1 + cos(theta) d(theta)=

integral sin(theta)sin(theta)= -sin(theta)cos(theta) + integral 1 d(theta) + integral cos(theta) d (theta)=

integral sin(theta)sin(theta)= -sin(theta)cos(theta) + theta - sin(theta) =

-sin(theta)cos(theta) -(theta)sin(theta) + C

Very close, but your trig identity isn't quite right:

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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18:47:06

what is the requested antiderivative?

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RESPONSE -->

antiderivative of sin^2 x=

integral sin(theta)sin(theta) = -sin(theta)cos(theta) - integral - cos(theta)cos(theta)=

cos^2 = 1 + cosx so,

integral sin(theta)sin(theta)= -sin(theta)cos(theta) + integral 1 + cos(theta) d(theta)=

integral sin(theta)sin(theta)= -sin(theta)cos(theta) + integral 1 d(theta) + integral cos(theta) d (theta)=

integral sin(theta)sin(theta)= -sin(theta)cos(theta) + theta - sin(theta) =

-sin(theta)cos(theta) -(theta)sin(theta) + C

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18:48:43

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

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RESPONSE -->

First I used f = sin(theta), f ' = cos(theta), g ' = sin(theta), g = -cos(theta)

Then I substituted cos^2 = 1 + cosx so,

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20:15:39

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

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RESPONSE -->

f= (2 + 3t)^^1/2

f ' = 1/2 (2 +3t)^ -1/2

g' = t + 2

g= 1/2 t^2 + 2t

integral of (t+2) `sqrt(2+3t) dt = (2 + 3t)^1/2 * 1/2t^2 + 2t - integral of 1/2 (2 + 3t)^ -1/2 * t + 2

integral of (t+2) `sqrt(2+3t) dt = t^2 (2 + 3t)^1/2 - 1/2 integral of (2 + 3t)^ -1/2 * t dt + integral of 2 dt

Now I solve for integral of (2 + 3t)^ -1/2 * t dt

f= (2 + 3t)^ 1/2

f ' = 1/2 (2 + 3t)^ -1/2

g' = t

g = 2t

integral of (2 + 3t)^ -1/2 * t dt= (2 + 3t)^1/2 * 2t - integral of 1/2 (2 + 3t) ^ -1/2 * t + 2

= 2t (2 + 3t)^1/2 - integral of t(2 + 3t)^ -1/2

integral of (2 + 3t)^ -1/2 * t dt= integral of 2t dt

Now I substitute:

f = (2 + 3t) ^ -1/2

f' = -1/2 (2 + 3t) ^ -3/2

g' = t

g= 2t

integral of (2 + 3t)^ -1/2 * t dt = (2 + 3t)^ -1/2 * 2t - integral of -1/2(2 + 3t)^ -3/2 * 2t

= 2t (2 + 3t) ^ - 1/2 + integral of t(2 + 3t) ^ -3/2

I can move the part on the right to the left and multiply both sides by (2 + 3t) in order to get the original function.

integral of (2 + 3t)(2 + 3t)^ -3/2 t dt = 2t (2 + 3t)^ -1/2 (2 + 3t)

=2t (2 + 3t) ^ 1/2 + integral of 2 dt

=2t (2 + 3t) ^ 1/2 + 2t

I put this back into the original function

integral of (t+2) `sqrt(2+3t) dt = t^2 (2 + 3t)^1/2 - 1/2 * 2t (2 + 3t) ^ 1/2 + 2t

= 3t - t^2(2 + 3t) + C

Again, good try. See the following;

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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20:15:49

what is the requested antiderivative?

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RESPONSE -->

= 3t - t^2(2 + 3t) + C

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20:15:53

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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20:38:18

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

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RESPONSE -->

f = cosx^3

f' = -3sinx^3

The derivative of cos(x^3) is -3 x^2 sin(x^3).

You need to remain aware of the chain rule.

g' = x^5

g= 1/6x^6

integral of x^5 cos(x^3) dx= cosx^3 * 1/6x^6 - integral of -3sinx^3* 1/6x^6

= 1/6x^6 * cosx^3 + 1/2 integral of x^6sinx^3

then I substitute:

f = sinx^3

f' = 3cosx^3

g ' = x^6

g= 1/7x^7

integral of x^6sinx^3= sinx^3 - 3/7 integral of x^7 cosx^3

I can take the part on the right to the left and subtract x to get the original.

integral of x^7 cosx^3 - x dx = -1/7x^7 * sinx^3 - 3/7 - x

Then I add this to the original equation.

integral of x^5 cos(x^3) dx= 1/6x^6 * cosx^3 + 1/2 - 1/7x^7 * sinx^3 - 3/7 - x

You made a fundamental error in taking your derivative of cos(x^3).

The derivative of your result does not equal x^5 cos(x^3). You need to begin taking the derivatives of your results, in order to check your work and discover errors of this nature.

It usually takes some trial and error to get this one:

We could try u = x^5, v ' = cos(x^3), but we can't integrate cos(x^3) to get an expression for v.

We could try u = cos(x^3) and v' = x^5. We would get u ' = -3x^2 cos(x^3) and v = x^6 / 6. We would end up having to integrate v u ' = -x^8 / 18 cos(x^3), and that's worse than what we started with.

We could try u = x^4 and v ' = x cos(x^3), or u = x^3 and v ' = x^2 cos(x^3), or u = x^2 and v ' = x^3 cos(x^3), etc..

The combination that works is the one for which we can find an antiderivative of v '. That turns out to be the following:

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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20:39:15

what is the requested antiderivative?

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RESPONSE -->

1/6x^6 * cosx^3 + 1/2 - 1/7x^7 * sinx^3 - 3/7 - x + C

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20:39:19

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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20:43:55

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 -->

I'm not sure how to calculate this since f is twice differentiable.

You don't need to know the specific function. 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.

STUDENT COMMENT: it seems awkward that the area is negative, so I believe that something is mixed up, but I have looked over it, and I`m not sure what exactly needs to be corrected

INSTRUCTOR RESPONSE: the integral isn't really the area. If the function is negative then the integral over a positive interval will be the negative of the area.

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20:43:57

What is the value of the requested integral?

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RESPONSE -->

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20:44:00

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

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RESPONSE -->

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20:44:05

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

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RESPONSE -->

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20:44:51

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

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RESPONSE -->

I think my main problem in integration by parts is knowing when to pull out a fraction and when to combine them.

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Be sure to see my notes.

Your basic approach is fine.

However you are making some very fundamental errors in taking derivatives, usually by neglecting the chain rule. Note also that you need to take the derivatives of your integrals in order to check your work.