Assignment 4

course Mth 174

?q?x?€?????y?v??assignment #004?????S??B+???Physics II

02-08-2009

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08:06:42

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

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RESPONSE -->

The integral of sin^2x=

sinx * sinx =

sinx = u -cosx = u'

sinx = v -consx= v'

the integral of sinxcosx + the integral of cos^2x =

the integral of sinxcosx + the integral of cosxsinx + the integral of sin^2x=

the integral of sinxcosx + integral cosxsinx + ingeral of 1 dx + integral cos^2dx

2 integral sinxcosx + integral cosxsinx + integral 1dx = sinxcosx + cosxsinx + x + C=

1/2sinxcosx +cosxsinx + x + C

Very good solution.

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.

 

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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08:06:47

what is the requested antiderivative?

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RESPONSE -->

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08:07:50

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

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RESPONSE -->

I used

u = sinx u'= cosx

v= sinx v' = cos x

and then did the same for cos^2x

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08:20:44

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

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RESPONSE -->

u= 2 + 3t du = 3 dx

x = u-2/3 dx= 1/3 du

1/3 integral of ((u-2)/3 + 2) squrt u * du

1/3 integral of (((2 + 3t) -2)/3 +2 squrt 2 + 3t * 3

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

1/2 integral of ((2 + 3t)/3) (2 +3t)^(1/2) + 13/3 + C

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08:20:50

what is the requested antiderivative?

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RESPONSE -->

u= 2 + 3t du = 3 dx

x = u-2/3 dx= 1/3 du

1/3 integral of ((u-2)/3 + 2) squrt u * du

1/3 integral of (((2 + 3t) -2)/3 +2 squrt 2 + 3t * 3

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

1/2 integral of ((2 + 3t)/3) (2 +3t)^(1/2) + 13/3 + C

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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08:21:15

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

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RESPONSE -->

i used

u= 2 + 3t du = 3 dx

x = u-2/3 dx= 1/3 du

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08:33:35

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

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RESPONSE -->

integral of x^5 cos(x^3)=

(1/3) x^3 sinx^3 + (1/3) cosx + C

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08:33:41

what is the requested antiderivative?

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RESPONSE -->

integral of x^5 cos(x^3)=

(1/3) x^3 sinx^3 + (1/3) cosx + C

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08:36:38

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

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RESPONSE -->

I factored out 1/3 and broke the rest of the equation into parts. I used

u= x^5 u' = 5x^4

v= cosx^3 v' = -sinx^3

The derivative of cos(x^3) is not - sin(x^3), it is - 3 x^2 sin(x^3).

However you have to get u and v ' from the integrand.

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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08:41:54

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 understood in where the book explained how to do integration by parts twice, but I don't know where to start when given f(0) =6, f(1) = 5 and f'(1)=2

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.

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08:41:57

What is the value of the requested integral?

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RESPONSE -->

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08:42:02

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

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RESPONSE -->

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08:42:38

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

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RESPONSE -->

I did not understand how to do #52.

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08:42:43

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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You need to be careful to use the chain rule when you are taking the derivative of a composite function. For example, as noted above, the derivative of cos(x^3) is not - sin(x^3).

In general any antiderivative you find should be checked by taking its derivative. If you consistently do so you will become much more aware of, and learn to avoid, some of the common pitfalls in the process of symbolic integration.

&#Let me know if you have questions. &#