#$&* course Mth 174 Questions from text assignment:
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used integration by parts to get the answer -sinx(cosx) - int(cos^2 x). confidence rating #$&*: 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(cos x (-cos x)) 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 COMMEN-T: 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 was on the right track but forgot that cos^x= 1 - sin^x. I should have then completed the problem to get -1/2 (sinx * cosx) + x/2 + C. ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Section 7.2 Problem 4 problem 7.2.4 (previously 7.2.16 was 7.3.18) antiderivative of (t+2) `sqrt(2+3t) **** what is the requested antiderivative? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using integration by parts, I got the answer: (3t+2)^(3/2) [ 30 (t+2) - 4(3t+2) ] / 135
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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): ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Section 7.2 Problem 8 **** problem 7.2.8 (previously 7.2.27 was 7.3.12) antiderivative of x^5 cos(x^3) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I could not figure out what to use for u and v’ for to continue in the integration by parts method. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: It usually takes some trial and error to get this one: • We could try u = x^5, v ' = cos(x^3), but we can't integrate cos(x^3) to get an expression for v. • We could try u = cos(x^3) and v' = x^5. We would get u ' = -3x^2 cos(x^3) and v = x^6 / 6. We would end up having to integrate v u ' = -x^8 / 18 cos(x^3), and that's worse than what we started with. • We could try u = x^4 and v ' = x cos(x^3), or u = x^3 and v ' = x^2 cos(x^3), or u = x^2 and v ' = x^3 cos(x^3), etc.. The combination that works is the one for which we can find an antiderivative of v '. That turns out to be the following: Let u = x^3, v' = x^2 cos(x^3). Then u' = 3 x^2 and v = 1/3 sin(x^3) so you have 1/3 * x^3 sin(x^3) - 1/3 * int(3 x^2 sin(x^3) dx). Now let u = x^3 so du/dx = 3x^2. You get 1/3 * x^3 sin(x^3) - 1/3 * int( sin u du ) = 1/3 (x^3 sin(x^3) + cos u ) = 1/3 ( x^3 sin(x^3) + cos(x^3) ). It's pretty neat the way this one works out, but you have to try a lot of u and v combinations before you come across the right one. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): We use x^3 for u and x^2cos(x^3) for v’. By plugging in the values for u, u’, v, and v’, we get 1/3 * x^3 sin(x^3) - 1/3 * int( sin u du ) = 1/3 (x^3 sin(x^3) + cos u ) = 1/3 ( x^3 sin(x^3) + cos(x^3) ). ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: **** What substitution, breakdown into parts and/or other tricks did you use to obtain the antiderivative?
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I basically used trial and error to find the antiderivative. confidence rating #$&*: 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): ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: Section 7.2 Problem 13 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? I used integration by parts to get u=x, v’=f’”(x), u’=1, and v=f’(x). I evaluated to get 1* f’(1) - (f(1) - f(0)) = 2+6-5 = 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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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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" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Section 7.2 Problem 13 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? I used integration by parts to get u=x, v’=f’”(x), u’=1, and v=f’(x). I evaluated to get 1* f’(1) - (f(1) - f(0)) = 2+6-5 = 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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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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" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!