# OpenLab Assignment #5: Exam 2 review and WeBWorK update

OpenLab Assignment (Due next Tuesday evening, March 16th).  Your second exam will take place next week.  The review sheet and answer key have been posted.  In the interest of making this assignment a help (instead of an extra chore), I’m asking you to post a comment related to the second exam.  Any one of the following types of comments will earn you full points:

• A request for help on a question.  It must be more specific than just “how do I do this problem?”  Let us know where you’re stuck (and what you’ve done so far).  If you don’t know how to get started on a problem, you can say so — but tell us a little about why (is it different from problems we did in class?).  Your post must use correct $\LaTeX$ notation!
• A general question about a topic that we’ve been studying.  Is there some part of the course that’s been bothering you?  A type of problem that you just don’t get?  Something you keep trying but always get wrong?  Post a question here – maybe someone can help.
• An answer to someone else’s question.  You don’t have to give every step or every detail, but you should provide enough information to help them along.  Your post must use correct $\LaTeX$ notation!
• A helpful comment or suggestion about a topic that we’ve been studying.  Do you have a trick for solving certain problems?  A neat way of remembering a formula?  Some thoughts about what makes certain problems hard, and suggestions for solving them? Post it!

WeBWorK Update:  As you all know, WeBWorK access has been terrible this week!  Our hardworking system administrator is trying to figure out what’s going wrong, but until the difficulties get worked out I am making the following changes:

1. WeBWorK assignment #1 will be extended for 2 weeks, and is now due on March 20th (this update will be made on the WeBWorK servers as soon as I can get access to them…).
2. There will be no new WeBWork assigned until the current difficulties are resolved.  We’ll pick up again once the system is working more reliably.
3. Some of you may have noticed Assignment #2 in the WeBWorK system.  This will be removed shortly — you don’t have to worry about it.

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### 139 Responses to OpenLab Assignment #5: Exam 2 review and WeBWorK update

1. I don’t understand how to integrate
$\int (e^x)^2 dx$
from problem 2

• Jonas Reitz says:

There have been a number of questions about this (see below). debitcard has it almost right in his response below — first, simplify the function to obtain:

$\int \pi e^{2x} dx$

next, use a substitution $u=2x$.

(note: since the real complication here is the “2” in front of the “x”, power users may be able to integrate this without substitution by making a guess as to the antiderivative and then making an adjustment by dividing by an appropriate constant — if this sounds confusing, just do the substitution).

Write back if you’re still stuck on this one.
Mr. Reitz

• $\int_0^1 (e^x)^2 dx \rightarrow (e^x)^2 = e^x \cdot e^x = e^(x+x) = e^2x$

Now that we’ve simplified our expression properly, apply our function to our formula:
$\int \pi \cdot (f(x))^2 dx \rightarrow \int_0^1 \pi \cdot e^2x dx \rightarrow \pi \int_0^1 e^2x dx$

If you don’t want to use Mr. Reitz’s “handwaving” method, where you work some magic on the integral and it suddenly transforms into your answer, your problem is now set and simplfied for U-Substitution:

$u=2x , du = 2 dx , \frac{du}{2} = dx \rightarrow \frac{\pi}{2} \cdot \int_{0u}^{1u} e^u du \rightarrow \frac{\pi}{2} \cdot ( \left. e^u \right|_{0u}^{1u} ) \rightarrow \frac{\pi}{2} \cdot ( \left. e^2x \right|_0^1 ) = \frac{\pi}{2} \cdot (e^{2x} - e^0) = \frac{\pi}{2} \cdot ( e^{2x } - 1 )$

If anyone was confused, I hope this helps

• I apologize. Those first two lines of math were supposed to appear as such:

Line 1: $\int_0^1 (e^x)^2 dx \rightarrow (e^x)^2 = e^x \cdot e^x = e^{(x+x)} = e^{2x}$

Line 2: $\int \pi \cdot (f(x))^2 dx \rightarrow \int_0^1 \pi \cdot e^{2x} dx \rightarrow \pi \int_0^1 e^{2x} dx$

• And to rectify one last mistake, the integrated form of the function is supposed to appear as $\frac{\pi}{2} ( \left. e^{2x} \right|_{0}^{1} )$ , instead of: $\frac{\pi}{2} ( \left. e^2x \right|_{0}^{1}$

• And the final answer should appear as: $\frac {\pi}{2} (e^2 - 1)$ not as: $\frac{\pi}{2} ( e^{2x} -1 )$

Again, I apologize.

2. theozeng says:

I’m guessing multiply e^x with e^x and integrate that.

3. debitcard says:

I’m not sure that problem can be integrated.

Here’s my approach so far

1) Multiply the inside (e^x) by the exponent (2) = e^2x
2) Take antiderivative
Note: e^2x=e^2x
To integrate e^2x, you add 1 to the exponent (2x) turns to 2x^2, then you divide by the new exponent (2)

so, x^2?

since it is a indefinite integral, attach a +C for constant.

• mendozak says:

I am also currently stuck at $\int_0^1 {\pi} e^{2x} dx$

• stan says:

This is my approach:
I did substitution U=2*x dU=2dx dx=dU/2, so we get

$\int_0^{\1} \frac{\pi*\e^u}{2} du$ =
(Pi/2)*e^u on the interval [0,1]

Stanislav Podolski MAT 1575 -6638

• stan says:

This is the part that did not parse
\int_0^1 \frac{{\Pi}*e^u}{2} du

• $\int_0^1 \frac{\pi \times e^u}{2} du$ is the formula that you were looking for that didn’t Parse.

• To clear that up: it’s written as such:

&#36 latex \int_0^1 \frac{\pi \cdot e^u}{2} du &#36

4. jishan007 says:

I can’t solve the problem 11(a) from our review sheet.
I assumed x = 6 sinθ. But what I am getting is
∫ (dθ/6(sinθ)^2 cosθ). then I am stuck.
Is there anyone who got the correct answer?

5. debitcard says:

u=x
a=6

a^2-u^2
u=asinθ

Yes, x=6sin (θ)
dx =6cos(θ) du

sinθ=x/6
θ=sin^-1(x/6)

I skip a few steps but,
Draw out reference triangle, substitute appropriately and simplify. You’ll wind up with

∫csc^2(u) /36 du

bring 1/36 into front = 1/36∫csc(u) du

The integral of csc^2 (u)= -cot(u)

I’ll let you work out the rest.

6. mmiltz says:

This question is based on the example we did in class relating to finding the area of a circle using calculus . I’m having trouble understanding why we let x=sin(θ), not cos(θ)?

• Jonas Reitz says:

Hi Melissa — in fact, it doesn’t matter: we could just as easily have used cosine instead of sine. All we really want is to be able to simplify the part under the radical using (some variation of) the identity $\sin^2 x + \cos^2 x = 1$.
Mr. Reitz

7. Karen L. says:

For problem 9e, I keep getting the solution as 1/2. Just wondering if my calculations are wrong or the answer key is wrong?

• Jonas Reitz says:

Hi Karen,
You’re absolutely right, the answer is $\frac{1}{2}$. Nice work — I’ll update the answer key on the OpenLab shortly.
Mr. Reitz

• Karen L. says:

I just found another problem. #11b i get $\frac{\sqrt{3}}{3} - \frac{\sqrt{2}}{2}$

• Karen L. says:

Never mind. I used x^2 instead of x^3

8. jeffreym says:

jeffrey mongal
mon/wed 4-5:40pm

Hey Prof,
Im stuck on 9g… i can’t seem to find a u and dv that actually works or make the problem doable. Please let me know what the eaisest way is to start the problem.
Thanks!

• stan says:

I solved it using Integration by parts, not sure if the answer is right though), but here it is:

After substituting U=secx dV=(secx)^2 dU=secx*tanx V=tanx , and solving it I got:

(tanx*secx-ln|secx+tanx|)/2 + C

Stanislav Podolski MAT 1575 -6638

9. koshygkoshy says:

Bibin Koshy MAT 1575 -6637
I am writing this on behalf of all those individuals who are still frustrated over the topic of cylinder shells and volume of solid. I was having the same problem but I can honestly say, after following a couple of rules, i was able to solve and see the volume of solids more efficiently. I want to share this mainly towards people who are still confused about whether integrate taken in respect to y or x in the cases of both methods.
The following are rules that I followed to cope up with the frustrations and achieve the volume of a solid more efficiently.
First thing I did before tackling a certain problem is, write the rules on a sheet of paper and simply follow it afterwards :
1)
In a washer method, when you are going about x axis, you always draw a perpendicular line from the region you are going about to the region enclosed by a certain curves and that is to see what the outer region and inner region is. When you usually see x as the axis you are going about, the limits of integration and all the functions must be in x format. However, when you see vertical axis such as x=-3 or x=2, you have to take into consideration, it is a vertical axis like y axis, therefore, everything must be in y format. In addition, something like y=2 or y=-7 is a horizontal axis or looks similar to the x axis, therefore, everything must be in the dx format. Another thing that one must take it to consideration, when you are going about the y axis, you have to convert given functions such as y=X^3 and convert it into a way that x is isolated, which in this case is x=f(y)or x=3sqroot(y). The thing to remember is, when you see a problem related to washer or disks, always look at the axis the problem is telling you to go about and integrate it accordingly.
2)
The cylindrical shell method:
Compared to the first method, washer or disk method, shell method is quite different, especially when integrating. In the washer method, all we had to do was look at whether x=3 or y=2 shows a vertical or horizontal axis, and from it, we could deduce that the integration must be done accordingly to the axis. However, for the shell method, that is not the same. When you want to find the volume of a solid using the shell method that is going about the x axis, people usually think the integration must be with respect to x, but when people see the clear picture, they will see that they are wrong and that everything is taken in the format of y. Likewise, when a problem says, ” find the volume of a solid using the shell method that is going about the y axis this time, everything must be taken in respect to dx or x.
*Like the first method, when you take integration with respect to y for the washer, always make sure the function such as y=sqrootof (x) is converted into a way you get the x isolated, like in this case x=y^2, so that proper integration can be done.

By knowing these key concepts of how they are different and similar , one can tackle these kind of problems related to volume more efficiently.

• Rich says:

That’s good observation. I followed the same rules and i get through that chapter as well.

• cthoma12 says:

i take back what i said in class LOL i read it i think i get it

10. Kedeshia says:

I am having trouble grasping the trigonometric integral problems. Can someone please explain some steps for me ?

• Tiff Ca$e says: Trigonometric integral problems are like regular integral problems except this time you change the “name” of the trig function to one that is equivalent helps simplify the problem. If you already know the anti- derivatives or derivatives, it will be a little easy to integrate. When it comes to trig functions with exponents, split it to where at least one new trig identity matches one you already know then proceed with the suitable integration method. Also, KNOW ALL TRIGONEMETRIC IDENTITIES, including degrees and equivalent fraction terms. Memorize almost everything important THEN try to tackle the problems. The point is to change it so integration could be easier than if you would’ve left it as is. For extra help outside the classroom… here’s a link- http://www.youtube.com/watch?v=lTqnlihOC4o&feature=fvwp&NR=1 Hope that helps! 11. Keyla says: For question number 2 on the review sheet, I am not sure how you get $pi/2 (e^2-1)$ I have $\int_0^{1} \e^u du$, where do I o from there? • cthoma12 says: $\frac{1}{2} \pi \int_0^{1} e^u du$ $\left. \right |_0^{1} = \frac{\pi}{2} (e^(2*1)-e^1))$ $u=2x ,\frac{1}{2}du=dx$ 12. Keyla says: idk whats wrong with latex.. I have pi/2 integral from 0 to 1, of function e^u • Jonas Reitz says: Hi Keyla — I think it just didn’t like the backslash in front of the e^u. 13. in the first quadrant bounded by the graph of y=4ln(3-x),the horizontal line y=6 and the vertical line x=2. find the volume of the solid generated when the graph is revolved about the horizontal line y=8. pi$latex\int_0^{\2}/ (6-4ln(3-x)^2)

ans = 168.80

• means pi is in outside ,then the intergral sign which goes 0 to 2 and in the (6-4ln(3-X)^2)

14. Becky Jett says:

i’m having the most difficult time trying to integrate $\int x^5lnx dx$. i think i used all possibilities for my u’s, du’s, and dv’s and i still keep getting the wrong answer. can anyone help me get on the right direction?

• Becky Jett says:

i would also want to know is how do you know you are getting the right or wrong answer when you are using integration by parts…?

• cthoma12 says:

$u = ln du=1/x v=x^5 dv = 1/6 x^6$

• Becky Jett says:

thats exactly what i got. thanks

15. jermin says:

im having a little problem with question #4
pi \int_-2{\2}\((4- x^2 +5)^2 – (x^2 -4 +5) from that i got;
(-x^2 + 9)^2 – (x^2 +1)^2 then foil;
(x^4 -18x^2 +81) – (x^4 +2x -1) after that i got this;
pi \int_-2{\2}\-20x^2 +80

• ximenam1123 says:

for this question this is the setup after this just integrate it and the bounds of the integral you get by x^2-4=4-x^2 and you should get x=-2,2
$\int_-2^{\2} \2pi(x+5)[(4-x^2)-(x^2-4)] dx$

• ximenam1123 says:

$\int_-2^{\2} \2\pi(x+5)[(4-x^2)-(x^2-4)] dx$

• ximenam1123 says:

$\int_(-2)^{\2} \2\pi(x+5)[(4-x^2)-(x^2-4)] dx$

• ximenam1123 says:

Prof.Reitz am putting the formula in latex but doesnt work \int_(-2)^{\2} \2\pi(x+5)[(4-x^2)-(x^2-4)] dx
am adding latex and $at the end but cant work it out. • Jonas Reitz says: Hi Ximena — two things: in the bounds of the integral, use curly brackets instead of parentheses around the -2, and later on don’t put a backslash in front of the number 2 (occurs twice). Give that a shot. -Mr. Reitz • ximenam1123 says: $\int_{-2}^{2} 2\pi(x+5)[(4-x^2)-(x^2-4)] dx$ • Jonas Reitz says: Hi Ximena – nice job on the latex! Be careful — this problem requires washers, not shells, since we are rotating around the horizontal line y=-5. This is going to affect your integral. 16. jermin says: do we have to know the trigonometric substitution by heart? • Jonas Reitz says: Unfortunately, yes! (but just the first two columns of the table – the function, and the corresponding substitution) 17. igorekk132 says: Igor. M. Heyy). In problem 8c I cant figure out how to integrate. Do I use technique of integration by parts? Where would I start ? $\int x^3e^xdx$ • cthoma12 says: integration by parts twice (or three times) keep your dv= to e^x and u=x^3 dv stays the same while u will change 18. ximenam1123 says: Hi Prof.Reitz How should I start question 11b . 19. jesus22 says: Hi Prof Reitz. i don’t understand how to do problem 10 in the review. • ximenam1123 says: For problem 10 x^2+y^2=9 and the center is (0,0) $\2 int_-3^{3} \sqrt{9-x^2} dx$ thats the set up for the problem • ximenam1123 says: $\int_-3^{3} \sqrt{9-x^2} dx$ • ximenam1123 says: $\int_-3^{\3} \sin x dx$ • ximenam1123 says: $\int_-3^{\3} \sqrt{9-x^2} dx$ • ximenam1123 says: the integral goes fro -3 to 3 20. theozeng says: I have a problem on how to start with problem 8C. $\int x^3*e^x dx$ Should I let u = x^3? or e^x? • tonymei999 says: well for this problem you have to use integration by parts three times. the first case U will equal $\x^3$ the second case U will equal to $\x^2$ and the 3rd case U will equal $\x$ Because this is a rather long problem pay attention to your constants and which integral each constant is a reference to. • bettygeorge says: $\int x^3e^x= x^3e^x-$latex \int x^33xdx u=x^3 dv=e^X$latex \int x^3e^X= x^3e^x-3x^2 3xdx
=e^x(x^3-3x^2+6x-6)

21. tonymei999 says:

well for this problem you have to use integration by parts three times. the first case U will equal $x^3$ the second case U will equal to $x^2$ and the 3rd case U will equal $x$ Because this is a rather long problem pay attention to your constants and which integral each constant is a reference to.

22. When I study a particular problem I like stop write the steps I have to follow to complete the problem as I work on it. It is sort of time consuming but it helps me to remember what I have to do to get a problem done successfully.

• Kedeshia says:

I do the same thing !

• cthoma12 says:

(even nerdier) i use a white board since my handwriting is equivalent to a five year olds, when things are bigger it helps me see them better and i can draw; clear cut lines and notes to myself

23. mrcpotter says:

I just wanted to share a weird way that i remember the integration by parts formula; although it isn’t the most difficult formula to remember…I just say to myself before a I.B.Parts question

“You are so DeVious for wearing UV sunglasses to protect from the Negative voodu”

u dv = uv – vdu
You devious UV (negative sign) Voodu

You just have to remember to put the integral at the beginning and after the negative so u will have the proper form below:
$\int udv = uv- \int vdu$

24. Joshua Ruiz says:

I’m still slightly confused by the basic concept of Cos2x… I can’t seem to remember how to maneuver it $latex \intcos{2x} • Joshua Ruiz says: $\int \cos{2x}$ • Jonas Reitz says: Hi Joshua, When the only thing in the way is a constant in front of x (like “2x”), try substitution: $u=2x$ Good luck, Mr. Reitz 25. KHonda says: problem 9) G$ latex \int sec x^3 dx $i used u= secx du=tanx secx secx^2* secx (tanx^2+1)*secx secxtan^2x+$ latex \int secx dx $26. KHonda says: problem 9) G $\int sec x^3 dx$ i used u= secx du=tanx secx secx^2* secx (tanx^2+1)*secx secxtan^2x+ $\int secx dx$ 27. lance12 says: I’m having slight difficulty with 9d. off the Review sheet: Any suggestions? please and thanks $\int sec^7x tan^3x dx$ • cthoma12 says: i just finished that one take out a sec*tan so that it’s now sec^6tan^2 tan*secdx then use u=sec du= sec*tan dx 🙂 (don’t forget the trig indent for tan^2) 28. cyeung1 says: 11B$ latex\int_sqrt[2]^{2}\frac{1}{x^2sqrt[x^2-1]$why is$latex\sqrt{2} is equal to sec Theta ?

• cyeung1 says:

$latex\int_sqrt[2]^{2}\frac{1}{x^2sqrt[x^2-1]$

$latex\sqrt{2}$

• cyeung1 says:

$\int_sqrt{2}^{\2}\frac{1}{x^2sqrt{x^2-1}$

$\sqrt{2}$

-don’t work.

29. Fengming08 says:

For question #8c:
I got $\= e^x (x^3-3 x^2-6 x-7)+c$
I I used $\u =x^3$ $\dv =e^x dx$
$\du =3 x^2 dx$ $\v= e^x$
I got $\int x^3 e^x dx = x^3 e^x - \int 3 x^2 e^x dx$
I used the intergration by parts again. $\u = 3 x^2$ $\dv =e^x dx$
$\du =6 x dx$ $\v= e^x$
I got $\int x^3 e^x dx = x^3 e^x - 3 x^2 e^x - \int 6 x e^x dx$

intergration by parts again: $\u = 6$ $\dv =e^x dx$
$\du = dx$ $\v= e^x$
$\= x^3 e^x - 3 x^2 e^x - 6 x e^x - 6 e^x - \int e^x dx$
$\= e^x (x^3-3 x^2-6 x-6-1)+c$

30. mendozak says:

For problem #6, When you do u-substitution for:

$latez \int_0^1 2{\pi}x \sin({{\pi}x^2})dx$

and get:

u = ${\pi}x^2$
du = $2{\pi}xdx$

I understand that you must then change bounds of the interval to

$\int_0^{/pi}$

and I understand how to do it, but I don’t understand why.

Can someone explain it to me?

• mendozak says:

Fixed.

For problem #6, When you do u-substitution for:

$\int_0^1 2{\pi}x \sin({{\pi}x^2})dx$

and get:

u = {\pi}x^2
du = 2{\pi}xdx

I understand that you must then change bounds of the interval to

\int_0^{pi}

and I understand how to do it, but I don’t understand why.

Can someone explain it to me?

• mendozak says:

FINAL revision. (They should REALLY implement an “edit post” feature.)

For problem #6, When you do u-substitution for:

$\int_0^1 2{\pi}x \sin({{\pi}x^2})dx$

and get:

u = ${\pi}x^2$
du = $2{\pi}xdx$

I understand that you must then change bounds of the interval to

$\int_0^{pi}$

and I understand how to do it, but I don’t understand why.

Can someone explain it to me?

• Tiff Ca$e says: After doing the substitution, u is the variable of integration, not x.. so the limits have to been put in terms of u in respect to x…. or you cannot finish the problem since you changed to u’s but left the integrals with x. I hope I kind of cleared it up a little. (Math is hard to explain in words for me) : ) • mendozak says: That makes sense. Thanks a lot! I have a hard time working arbitrarily. I have to know why I’m doing something before I start doing it. 31. Hi, can anyone help me with question 8D from the review sheet? Using integration by parts i have u=sinx and dv=e^x but each time i finish integrating i am left with what i started with, so i am going in circles. What is the trick? What can i do? • Bettyann John-George says: \int_e^x sinxdx u=e^x du=e^x dv=sinx v=-cosx \int_e^x sinx=e^x-cosx-\int_-cosx e^x dx \int_ e^x sinx=-e^xcosx+\int_e^x cosx dx 2\int_ e^x sinx=-e^x cosx+e^x sinx \int_ e^x sinx=1/2 e^x(sinx-cosx)+C • Bettyann John-George says: I could not get the integral sign to work but that’s how I did (8)d. • cthoma12 says: $\int \sin(x) e^x=somenumbers -\int \sin e^x$ • cthoma12 says: you will see this eventually then just divide the two so it becomes $\frac {e^x(\sin(x) -cos(x))}{2}$ 32. Hi, i am having problem solving question 11c on review. 33. Hi, i am having problem solving question 11c on review. 34. Drake says: Just ran into some tips that I found helpful regarding to when to use the washer or shell method. When the dependent variable and the axis parallel to the axis of rotation are the same, use the washer method. When the dependent variable and the axis parallel to the axis of rotation aren’t the same, use the shell method. Hopefully this will help someone else. Good luck to everyone tomorrow! • Tiff Ca$e says:

Wow…….. it really did . Thanks a lot!

35. Fengming08 says:

For question #8c:
I got $latex\ e^x(x^3-3 x^2-6 x-7)+C$
I I used $latex\ u=x^3$ $latex\ dv=e^x dx$
$latex\ du=3 x^2 dx$ $latex\ v=e^x$
I got $latex\ x^3 e^x-\int e^x 3 x^2 dx$
I used the intergration by parts again.
$latex\ u=3 x^2$ $latex\ dv=e^x dx$
$latex\ du=6 x dx$ $latex\ v=e^x$
I got $latex\ x^3 e^x-3 x^2 e^x-\int 6 x e^x dx$
intergration by parts again: $latex\ u=6$ $latex\ dv=e^x dx$
$latex\ du=dx$ $latex\ v=e^x$
$latex\ x^3 e^x-3 x^2 e^x-6 x e^x-6 e^x-\int e^x dx$
$latex\ e^x(x^3-3 x^2-6 x-6-1)+C$

36. Fengming08 says:

For question #8c:
I got
$latex\ e^x(x^3-3 x^2-6 x-7)+C$
I I used
$latex\ u=x^3$
$latex\ du=3 x^2 dx$
$latex\ dv=e^x dx$
$latex\ v=e^x$
$latex\ x^3 e^x-\int e^x 3 x^2 dx$
I used the intergration by parts again.
$latex\ u=3 x^2$
$latex\ du=6 x dx$
$latex\ dv=e^x dx$
$latex\ v=e^x$
$latex\ x^3 e^x-3 x^2 e^x-\int 6 x e^x dx$
intergration by parts again:
$latex\ u=6$
$latex\ du=dx$
$latex\ dv=e^x dx$
$latex\ v=e^x$
$latex\ x^3 e^x-3 x^2 e^x-6 x e^x-6 e^x-\int e^x dx$
$latex\ e^x(x^3-3 x^2-6 x-6-1)+C$

• Fengming08 says:

I don’t know why it does’t work???

• cthoma12 says:

$latex\ e^x(x^3-3 x^2-6 x-7)+C$
I I used
$latex\ u=x^3$
$latex\ du=3 x^2 dx$
$latex\ dv=e^x dx$
$latex\ v=e^x$
$latex\ x^3 e^x-\int e^x 3 x^2 dx$
I used the intergration by parts again.
$latex\ u=3 x^2$
$latex\ du=6 x dx$
$latex\ dv=e^x dx$
$latex\ v=e^x$
$latex\ x^3 e^x-3 x^2 e^x-\int 6 x e^x dx$
intergration by parts again:
$latex\ u=6$
$latex\ du=dx$
$latex\ dv=e^x dx$
$latex\ v=e^x$
$latex\ x^3 e^x-3 x^2 e^x-6 x e^x-6 e^x-\int e^x dx$
$latex\ e^x(x^3-3 x^2-6 x-6-1)+C$

• cthoma12 says:

Z_Z
Z_z
$e^x(x^3-3 x^2-6 x-7)+C$
I I used
$u=x^3$
$du=3 x^2 dx$
$dv=e^x dx$
$v=e^x$
$x^3 e^x-\int e^x 3 x^2 dx$
I used the intergration by parts again.
$u=3 x^2$
$du=6 x dx$
$dv=e^x dx$
$v=e^x$
$x^3 e^x-3 x^2 e^x-\int 6 x e^x dx$
intergration by parts again:
$u=6$
$du=dx$
$dv=e^x dx$
$v=e^x$
$x^3 e^x-3 x^2 e^x-6 x e^x-6 e^x-\int e^x dx$
$e^x(x^3-3 x^2-6 x-6-1)+C$

• cthoma12 says:

for the -1 i’m not sure where you got it from the last one int e^x is already factored out unless you put in du=1 but +C is there

• zbrowne says:

$latex\ u=6$
$latex\ du=dx$
$latex\ dv=e^x dx$
$latex\ v=e^x$
$latex\ x^3 e^x-3 x^2 e^x-6 x e^x-6 e^x-\int$ $latex\ e^x dx$
$latex\ e^x(x^3-3 x^2-6 x-6)+C$
It should just be the above answer, no (-1) after the -6 is necessary.

• Jonas Reitz says:

Hi guys — just get rid of the slash \ after the word latex.
-Mr. Reitz

37. Musaib says:

$\int secx^3 dx$ I did this problem. However, i am getting 1/4 (sec x tan x).

38. endri says:

hi anyone can share the problem 9e from the review sheet?
which rule is it used?integration by parts?

39. vasquez says:

I think that a good exam problem would be to find the volume of
y= 1/x , x=1, x=2, y=0 about the x-axis.

• endri says:

prof reitz doesn’t give easy problems like this, if a problem is not 2 pages long, then he probably wont give it in the exam ;p

40. Fengming08 says:

I have a question on #4.
How to figure out which fuction is the small radius and the large radius?

• If you’re still in the dark about that for #4, the trick to figuring out that one is pretty simple. Whichever function is further away from the line that you’re rotating around is the larger radius. So since $4 - x^2$ is further away from $y = -5$ , it’s the larger radius in the function. This concept works all the time.

• If that’s still not quite enough for you, you can use a more mathematical method and a less visual method, but they both come out to the same thing. Plug one constant value for x into both of your functions within the region that we’re integrating. You take those values and find the distance from the line around which you’re rotating and see which comes up bigger.

For example:

$x = 0 \rightarrow y = x^2 - 4 \rightarrow y = (0) -4 \rightarrow y = -4 ; y = 4 - x^2 \rightarrow y = 4 - 0 = 4$

To find the distance between that value in either function and the line that you’re rotating around:

For $y = x^2 - 4 \rightarrow y = -4$ :

$-4 - (-5) = 1$

For $y = 4 - x^2 \rightarrow y = 4$ :

$y 4 - (-5) = 9$

$4 - x^2$ is the function with the larger radius.

• Fengming08 says:

I was rotating around the line x=-5.

41. I have a question for the third question on the review sheet. We are supposed to integrate the region bounded by: $y = -2x^2 + 6 and y = 0$ In order to find the bounds for our integral, we use a system of equations, which should look like this: $-2x^2 +6 = 0 \rightarrow 2x^2 = 6 \rightarrow x^2 = 3 \rightarrow x = sqrt{x}$ If this is correct, then why are the bounds of integration [0,3] instead of $[0,\sqrt{3}]$ ? Did I go wrong somewhere, or is the answer in the key flawed?

• To clear that first part up a bit: $y = -2x^2 +6$ and $y = 0$

• Fengming08 says:

The function is $latex\ y=-2x^2+6x$ not $latex\ y=-2x^2+6$
so $latex\ 0=-2x^2+6x x(-2x+6)=0 x=0 -2x+6=0 x=3 • Thanks very much. Not sure why I didn’t see that one. • Fengming08 says: $\y = -2 x^2 + 6 x$ not $\y = -2 x^2 + 6$ so $= -2 x^2 + 6 x$ x(-2x+6)=0 x=0 -2x+6=0 x=3 42. mmiltz says: This isn’t specifically related to the homework, but I figured I’d throw it on here: Happy Pi Day! 43. takther2009 says: for number four i dont know what to do. there is an extra 20.. when i get ride of 20 it works but i dnt kno how i got the extra 20. 44. cthoma12 says: • Jonas Reitz says: Funny! I am extremely disturbed that I find myself nodding in agreement with many things she said… 45. Dear Professor, I am sorry I could not post this yesterday. I have a problem initially recognizing some problems from the onset, but I am getting better. My other problem is when I set up the equations; I am having some algebraic problems, especially sign distribution. I am trying very hard to slow myself down, as I work through the longer problems. 46. HI Prof, Quick Question – Are you going to provide us with the reduction formulas if need? Because we need to use the reduction formula for 9G. • Jonas Reitz says: It is technically possible to do 9g using only things we’ve covered (convert secants to tangents, integration by parts, and the integral of secant), BUT it’s a tricky one – I won’t put it on the exam. 47. avald1046 says: In case you are having trouble remembering integration by parts here’s an awesome tutorial!! http://www.youtube.com/watch?v=dqaDSlYdRcs 48. shamonie22 says: I dont know how to create a new topic but i need help on a webwork question. I’m asked to find the volume of a solid between the regions y=17 and y=x+72/x. I tried the shells method and used the formula$latex\int_8^{9}(x+3)(x+\frac{72}{x}-17)dx\$ I also tried washer with the formula $\int_8^{9}{\pi(17+3)^2-\pi(x+\frac{72}{x}+3)^2}dx$

49. shamonie22 says:

I dont know how to create a new topic but i need help on a webwork question. I’m asked to find the volume of a solid between the regions y=17 and y=x+72/x. I tried the shells method and used the formula $\int_8^{9}(x+3)(x+\frac{72}{x}-17)dx$ I also tried washer with the formula $\int_8^{9}{\pi(17+3)^2-\pi(x+\frac{72}{x}+3)^2}dx$

50. shamonie22 says:

I forgot to mention that it rotates around x= -3.

• shamonie22 says:

2pi should be in the first formula. That was an error.