Open query 4

course mth174

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Question: 7.2.3 (previously 7.2.12. (3d edition 7.2.11, 2d edition 7.3.12)) Give an antiderivative of sin^2 x

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Your solution:

Sin x = 1 – cos 2x / 2  ½ (x – sin 2x / 2) + c

Sin 2x  2 sin x cos x

Thus: ½ (x – sin x cos x) + c

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Given Solution: 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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Question: problem 7.2.4 (previously 7.2.16 was 7.3.18) antiderivative of (t+2) `sqrt(2+3t)

**** what is the requested antiderivative?

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Your solution:

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

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Given Solution: 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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Question:

**** problem 7.2.8 (previously 7.2.27 was 7.3.12) antiderivative of x^5 cos(x^3)

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Your solution:

U= x^3

V= 1/3 sin x^3

U’ = 3x^2

V’ = x^2 cos (x^3)

1/3 * x^3 sin x^3 – 1/3 * 3 x^2 sin x^3 dx

1/3 (x^3 sin x^3 + cos x^3)

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Given Solution:

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

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Your solution:

I tried u = x^4

u’= 4x^3

v= x cos (x^3)

v’ x^2 cos (x^3)

I believe I got mixed up on what to was to be separated and that’s where I messed up.

You're on the right track, but you need to break the integrand into u and v ', not into u and v.

Just a small change would do it: u = x^3 and v ' = x^2 cos(x^3) works, as shown in the given solution.

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Given Solution:

TYPICAL STUDENT COMMENT:

I tried several things:

v'=cos(x^3)

v=int of v'

u=x^5

u'=5x^4

I tried to figure out the int of cos(x^3), but I keep getting confused:

It becomes the int of 1/3cosudu/u^(1/3)

I feel like I`m going in circles with some of these.

INSTRUCTOR RESPONSE:

As noted in the given solution, it often takes some trial and error. With practice you learn what to look for.

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00:53:03

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Question: problem7.2.13 (previously 7.2.50 was 7.3.48) f(0)=6, f(1) = 5, f'(1) = 2; find int( x f'', x, 0, 1).

**** What is the value of the requested integral?

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Your solution:

f(0) = 6

f(1) = 5

f’(1) = 2

u= x

u’= 1

v= f’(x)

v’= f’’(x)

x f’(x) – f(x)

1 * f’(1) – [ f(1) – f(0) ]

 F’(1) – [ f(1) – f(0) ]

2 – [5 + 6]

2 – [-1]

= 3

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Given Solution:

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

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

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00:58:57

This was a very tedious assignment, but it will surely be a useful tool in computing areas over fixed integrals in the future. I do need more practice at these integrals, because I feel as if I`m going in circles on some of them. Any suggestions for proper techniques or hints on how to choose u and v? I have tried to look at how each variable would integrate the easiest, but I seem to make it look even more complex than it did at the beginning.

** you want to look at it that way, but sometimes you just have to try every possible combination. For x^5 cos(x^3) you can use

u = x^5, v' = cos(x^3), but you can't integrate v'. At this point you might see that you need an x^2 with the cos(x^3) and then you've got it, if you just plow ahead and trust your reasoning.

If you don't see it the next thing to try is logically u = x^4, v' = x cos(x^3). Doesn't work, but the next thing would be u = x^3, v' = x^2 cos(x^3) and you've got it if you work it through.

Of course there are more complicated combinations like u = x cos(x^3) and v' = x^4, but as you'll see if you work out a few such combinations, they usually give you an expression more complicated than the one you started with. **

This assignment was very time consuming because many of the problems had to be worked several times to achieve a

suitable answer. I will definitely need to practice doing more

** Integration technique does take a good deal of practice. There really aren't any shortcuts.

It's very important, of course, to always check your solutions by differentiating your antiderivatives. This helps greatly, both as a check and as a way to begin recognizing common patterns. **

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&#Good responses. See my notes and let me know if you have questions. &#

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