course MTH 174
7:03 p.m.
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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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: There are two antiderivatives for this function. Since this section is integration by parts, I will work it that way. Evaluating Int( sin^2 x dx) with u= sin x, u'= cos x; v'=sin x, v= -cos x. Using u*v- Int(u'*v) gives (sin x)(-cos x)- Int((cos x)(-cos x)). Using a familiar trig substitution gives -sin x cos x- Int(1 dx)-Int(sin^2 x dx). Evaluating the first integral makes this Int(sin^2 x dx)=-sin x cos x-x-Int(sin^2 x). Now we add the last integral to the left side of the equation and divide by the resulting 2 on the left side and we have the desired integral on the right: 1/2(x-(sin x * cos x)) + c.
confidence rating #$&* 3
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Given Solution: Good student solution:
The answer is -1/2 (sinx * cosx) + x/2 + C
I arrived at this using integration by parts:
u= sinx u' = cosx
v'= sinx v = -cosx
int(sin^2x)= sinx(-cosx) - int(cosx(-cosx))
int(sin^2x)= -sinx(cosx) +int(cos^2(x))
cos^2(x) = 1-sin^2(x) therefore
int(sin^2x)= -sinx(cosx) + int(1-sin^2(x))
int(sin^2x)= -sinx(cosx) + int(1) – int(sin^2(x))
2int(sin^2x)= -sinx(cosx) + int(1dx)
2int(sin^2x)= -sinx(cosx) + x
int(sin^2x)= -1/2 sinx(-cosx) + x/2
INSTRUCTOR COMMENT: This is the appropriate method to use in this section.
You could alternatively use trigonometric identities such as
sin^2(x) = (1 - cos(2x) ) / 2 and sin(2x) = 2 sin x cos x.
Solution by trigonometric identities:
sin^2(x) = (1 - cos(2x) ) / 2 so the antiderivative is
1/2 ( x - sin(2x) / 2 ) + c =
1/2 ( x - sin x cos x) + c.
note that sin(2x) = 2 sin x cos x.
Self-critique (if necessary): I only thought of sin^2 x= 1/2 - 1/2 cos 2x. I didn't see sin (2x)= 2 sin x cos x.
One correct answer is of course sufficient, and the two solutions are equivalent.
Self-critique rating #$&* 3
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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?
Your solution: The integral is: Int((t+2)(2+3t)^1/2 dt). Using u= t+2, u'= 1 dt; v'= (2+3t)^1/2, v= 2/9(2+3t)^3/2. With the parts equation we now have: ((t+2)(2/9(2+3t)^3/2) - Int( 2/9(2+3t)^3/2 dt). Evaluation of the integral leaves the desired antiderivative: ((t+2)(2/9(2+3t)^3/2) - 4/135(2+3t)^5/2 + c.
confidence rating #$&* 3
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Given Solution: If you use
u=t+2
u'=1
v'=(2+3t)^(1/2)
v=2/9 (3t+2)^(3/2)
then you get
2/9 (t+2) (3t+2)^(3/2) - integral( 2/9 (3t+2)^(3/2) dt ) or
2/9 (t+2) (3t+2)^(3/2) - 2 / (3 * 5/2 * 9) (3t+2)^(5/2) or
2/9 (t+2) (3t+2)^(3/2) - 4/135 (3t+2)^(5/2). Factoring out (3t + 2)^(3/2) you get
(3t+2)^(3/2) [ 2/9 (t+2) - 4/135 (3t+2) ] or
(3t+2)^(3/2) [ 30/135 (t+2) - 4/135 (3t+2) ] or
(3t+2)^(3/2) [ 30 (t+2) - 4(3t+2) ] / 135 which simplifies to
2( 9t + 26) ( 3t+2)^(3/2) / 135.
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Self-critique (if necessary): I didn't think to do the factoring to reduce the final equation.
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Self-critique rating #$&* 3
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Question:
**** problem 7.2.8 (previously 7.2.27 was 7.3.12) antiderivative of x^5 cos(x^3)
Your solution: Int(x^5 cos(x^3) dx). We can factor out an x^2 from the x^5 to make; Int(x^2 x^3 cos(x^3) dx). Using u= x^3, u'= 3x^2; v'= x^2 cos(x^3), v= 1/3 sin(x^3) we then evaluate: (x^3)(1/3 sin(x^3)) - Int(3x^2 sin(x^3) dx) and finally have: 1/3x^3 sin(x^3) - 1/3 cos(x^3) + c
confidence rating #$&*
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Given Solution:
It usually takes some trial and error to get this one:
• We could try u = x^5, v ' = cos(x^3), but we can't integrate cos(x^3) to get an expression for v.
• We could try u = cos(x^3) and v' = x^5. We would get u ' = -3x^2 cos(x^3) and v = x^6 / 6. We would end up having to integrate v u ' = -x^8 / 18 cos(x^3), and that's worse than what we started with.
• We could try u = x^4 and v ' = x cos(x^3), or u = x^3 and v ' = x^2 cos(x^3), or u = x^2 and v ' = x^3 cos(x^3), etc..
The combination that works is the one for which we can find an antiderivative of v '. That turns out to be the following:
Let u = x^3, v' = x^2 cos(x^3).
Then u' = 3 x^2 and v = 1/3 sin(x^3) so you have
1/3 * x^3 sin(x^3) - 1/3 * int(3 x^2 sin(x^3) dx).
Now let u = x^3 so du/dx = 3x^2. You get
1/3 * x^3 sin(x^3) - 1/3 * int( sin u du ) = 1/3 (x^3 sin(x^3) + cos u ) = 1/3 ( x^3 sin(x^3) + cos(x^3) ).
It's pretty neat the way this one works out, but you have to try a lot of u and v combinations before you come across the right one.
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Self-critique (if necessary): OK
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Self-critique rating #$&* 3
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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 started it like normal and when I was integrating cos x^3 to use as v I realized I needed an x^2 from somewhere and then I saw I could just factor it out of x^5.
confidence rating #$&* 3
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Given Solution:
TYPICAL STUDENT COMMENT:
I tried several things:
v'=cos(x^3)
v=int of v'
u=x^5
u'=5x^4
I tried to figure out the int of cos(x^3), but I keep getting confused:
It becomes the int of 1/3cosu du/u^(1/3)
I feel like I`m going in circles with some of these.
INSTRUCTOR RESPONSE:
As noted in the given solution, it often takes some trial and error. With practice you learn what to look for.
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Self-critique (if necessary): This one didn't give me as much trouble some of the others (I'm looking at you Mr. Arctan(4z)!).
Right. That one isn't bad if you get the integration by parts right, but until you get used to the trick it's sort of counterintuitive to let u = arctan(4z).
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Self-critique rating #$&* 3
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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), x, 0, 1).
**** What is the value of the requested integral?
Your solution: Using u = x, u' = dx; v'= f''(x), v= f'(x) gives x f'(x) - Int(f''(x) dx) and then x f'(x) - f(x). Then using the values given over the interval: (1*2 - 5) - (0 - 6)= 3.
confidence rating #$&* 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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My input: I was surprised at how long it took to work so few problems. When you have two substitutions its hard to see at first where your headed then trying to remember those trig identities, and keeping the u,u' and v', v choices consistent in the double substitutions all added up to make the section the most challenging yet.
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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 work. See my notes and let me know if you have questions. &#
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