Assignment 4

course Mth 174

This assignment was pretty tough. I will need to spend more time with substitution and integration by parts.

݄𗰴Ш䙺assignment #004

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Physics II

06-08-2008

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21:28:07

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

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RESPONSE -->

I used the inegration by parts method.

u = sin x

u' = cos x

v' = sin x

v = -cos x

int(sin x * sin x) = sin x * -cos x - int(cos x * -cos x)

int(sin^2 x) = sin x * -cos x - int(cos^2 x)

That was as far as I got.

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.

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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21:28:32

what is the requested antiderivative?

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RESPONSE -->

sin x * -cos x - int(cos^2 x)

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21:29:22

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 parts.

u = sin x

u' = cos x

v' = sin x

v = -cos x

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21:59:09

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

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RESPONSE -->

I am still really bad at substitution, which is what I tried to use to answer this problems.

w = 2 + 3t

w' = 3

t = (w-2) / 3

t' = 1/3

(w - 2) / 3 + 2 * sqrt(w) + C

Not a bad try, but substitution gets to be a mess on this problem. It works out easier if you integrate by parts.

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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21:59:11

what is the requested antiderivative?

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RESPONSE -->

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21:59:13

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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22:10:16

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

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RESPONSE -->

I used substitution again on this one.

w = x^3

w' = 3x^2

x = w^(1/3)

x' = (1/3) * w^(4/3)

int[w^(1/3)^(5) * cos^w]

int[w^(5/3) * cos^w]

(3/8)w^(8/3) * (sin^(w + 1)) / (w +1) + C

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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22:17:36

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 would guess that the int(x * f''(x)) would be (1/2)x^2 * f'(x) + C

x is a factor of the integrand, and you can't integrate it separately. In effect, the product rule does you in if you do that.

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.

When x = 1, (1/2) * 1 * 2 + C= 1 + C

Because when x = 1, f'(x) = 2

This is as far as I got

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22:17:38

What is the value of the requested integral?

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22:18:16

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

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RESPONSE -->

I used substitution. I know this chapter is about integration by parts, but I am still trying to be good at substitution

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22:18:19

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

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You need more practice with this, but you've got a good start. Follow my previous advice about doing a lot of problems, and feel free to submit questions on anything.