#$&* course mth 174 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Sin^2x=(1-cos(2x)/2 I separated this into 1/2-cos(2x)/2 this then becomes 1/2(x-sin2x)/2+c which simplifies to ½*(sinx*cosX)+c confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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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 am having some trouble with Integration by parts and a t this point I realize I need to look over it again. But I understand the alternative solution fairly well.
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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 was having trouble with the simplification of this problem. After looking at this for a I now see that the dt changes the power of the second component to (5?2) then I was hung up on the factoring but I now see that we can factor out the (3/2 ) out of the 5/2 because it will result in a value of 2/2 which is equal to 1 which can be omitted from the equation. Once I figured this out the rest of the problem appears to be fairly straight forward.
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I Had trouble how to finding the correct U and V values can you explain to me how to do this other than the trial and error approach that I am using. I went through the same process on paper as the solution I just never actually made it to the problem answer.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In most cases I try to start with substitution and if that proves not possible I then breakdown into parts. The problem is that frequently I do not see it initially then after seeing the solution it is clear. 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 1/3cosudu/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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am like the typical student comment in the fact that I am spending a lot of time in circles and in many cases abandonoong my work and restarting but I think that my big thing is I need more practice.
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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. ** #$&* "