Review for Third and Fourth parts (section 008)

Note from the instructor: I have put together the review exams for the third and fourth parts of the course, as we commented in class. Both tests were very good, although the fourth had many repeated questions, and some questions that did not belong in this part. I have changed the coefficients of all problems to prevent the use of answer keys.

  1. Compute the following indefinite integrals:
    1. \displaystyle{\int \frac{27 \ln t}{3t^2}\, dt}
    2. \displaystyle{\int \frac{4x^3}{13x^4-7}\, dx}
    3. \displaystyle{\int \frac{2x}{\sqrt{3x^2+1}}\, dx}
    4. \displaystyle{\int 2x e^{5x^2}\, dx}
    5. \displaystyle{\int (t-17) \sqrt{2t+3}\, dt}
    6. \displaystyle{\int 4x^2 \ln x\, dx}
    7. \displaystyle{\int \frac{x+2}{x^2+4x+23}\, dx}
    8. \displaystyle{\int 27t^2 e^{23t}\, dt}
    9. \displaystyle{\int (6x^2-8x)^{23} (6x-4)\, dx}
  2. Using the Fundamental Theorem of Calculus, evaluate the following definite integrals:
    1. \displaystyle{\int_0^{10} (7-x^2)^2\, dx}
    2. \displaystyle{\int_{1/2}^1 (3t^2+4t+3)\, dt}
  3. Find the area under the curve y=x^2 between x=1 and x=4.
  4. Compute the net area of the function f(x)=1-x^2 between x=-\sqrt{2} and x=\sqrt{2}.
  5. Find an antiderivative F(x) of f(x)=3x^2-13x+1 that satisfies F(1)=3
  6. A vintage toy was worth $33 in 1970 and its price increased at a rate of 19% per year.
    1. Find an equation of the value V of the toy in dollars, t years after 1970.
    2. Find the average cost of the toy over the period from 1970 to the year 2013.
    3. What year was the closest to the average cost of the toy?
  7. If the demand and supply are given by p=2001-22q and p=27q+26:
    1. Which one is the demand, and which one is the supply?
    2. Find consumer and producer surplus at equilibrium
  8. Find the area of the region bounded by the graphs of y=x+5, y=3x-3, and x=-2.
  9. Find the area of the region bounded by the graphs of x=y+3 and x=y^2 from y=-1 to y=2.
  10. (Note from the instructor: this problem is awesome. Challenging, but really good to test many of your algebraic skills) The yearly consumption of fruit in the US increased exponentially since 1900 at a continuous rate of 2.9% per year. The amount of fruit consumed in 1950 was 4.5 million pounds.
    1. Find a formula for the yearly consumption of fruit in millions of pounds, F, as a function of time t in years since 1900.
    2. Find the average yearly fruit consumption between 1950 and the year 2000.
    3. During what year was fruit consumption closest to average?
  11. The mall cops at Five Finger Discount Mall noticed an increase in department store theft after their local economy plummeted. In one particular store, they recorded the dollar amount of all items stolen from 2000 to 2010. The yearly amount of stolen items is listed below:

    \begin{tabular}{l|r|r|r|r|r|r|r|r|r|r}  & 2000 & 2001 & 2002 & 2003 & 2004 & 2005 & 2006 & 2007 & 2009 & 2010 \\ \hline \$ & 200 & 276 & 283 & 350 & 412 & 430 & 476 & 499 & 501 & 530 \end{tabular}

    Estimate the average dollar amount of items stolen from 2005 to 2010.

  12. If t is measured in days since June 1st, the inventory for an item in a warehouse is given by I(t)=5000 (0.8)^t. Find the average inventory in the warehouse during the 60 days after June 1st.
  1. Avery Ducharme
    April 29, 2013 at 6:17 pm

    Can someone please explain #1 letter h please

    • Katherine Stein
      May 1, 2013 at 7:36 pm

      Solve by integration by parts. The formula for it is ∫u*dv = u * v – ∫v * du.
      u=27t^2
      dv=e^23t
      du=54t
      v=e^23t/23

      Plug in those values into the equation above. (27t^2)(e^23t/23) – ∫(e^23t/23)(54t)
      Simplify everything after the integral sign before taking the integral of it.

  2. Juan-Carlos Valdez
    April 29, 2013 at 8:18 pm

    For parts a and c of #1 do you use integration by parts or substitution

    • Katherine Stein
      May 1, 2013 at 7:42 pm

      A is integration by parts and C can be solved by either method, if you bring the denominator up by making the exponent of the function -1/2. To solve it by substitution, let u=3x^2+1. Find the derivative of that, which is just 6x. You need to get the derivative of u to equal the du in the original function (2x) so you will need to divide by 3. You can notate that by putting 1/3 outside of the integral.

      Now you have this: (1/3)∫du/u. Taking the integral of du is just 1, and the integral of 1/u is ln(u). So after the integration is complete you are left with (1/3) * ln (u). Then you can plug your u back into that, and that’s the answer.

      (1/3)ln(3x^2+1)

  3. Ben Zeigler
    April 30, 2013 at 1:18 pm

    Does anyone know which rule is used to solve #1 a

    • Brooke Balentine
      April 30, 2013 at 10:45 pm

      If I’m right you use u and dv. I think that the answer comes out to be
      27lnt(-3^-1)-81t^-1 if I’m right.

  4. Lauren Bennett
    April 30, 2013 at 1:47 pm

    Can someone from group 3 explain how to work #1 part a and b?

    • Nicholas Long
      April 30, 2013 at 6:16 pm

      I’m not sure how to do part (a), but for (b)
      if you take the derivative of the bottom then you get 52x^3, and since the top is 4x^3, you need to take 13 and add 13. When you do this the answer is 1/13|ln13x^4-7|.

  5. Taylor Fary
    April 30, 2013 at 2:23 pm

    For number 3 do you just integrate x^2 and evaluate at 4 and 1 and then subtract?

    • Nicholas Long
      April 30, 2013 at 6:09 pm

      For number 3, you integrate x^2 and get 1/3x^3, and then you evaluate both 4 and 1 and then subtract.

    • Carly Gillet
      May 1, 2013 at 8:57 pm

      Yes, this is similar to a definite integral where your f(x) is x^2 and b=4 and a=1. So you end up subtracting 1/3(4)^3 – 1/3(1)^2.

  6. Brooke Balentine
    April 30, 2013 at 10:47 pm

    Does anyone know how to do #1 letter e and letter h. I’m stuck.

    • Nick Calcanes
      May 1, 2013 at 3:31 pm

      For both you would have to do integration by parts. I don’t know if you have the formula for it in your notes but it’s ∫udv=uv-∫ vdu

    • Nicholas Long
      May 1, 2013 at 3:33 pm

      The way to do e is to use the u, du, and v. Your u= t-17, du= dv, and your
      v= 1/3(2t+3)^3/2.
      Then you plug those numbers into the rule (A11), the integration by parts.
      And if I did my math right your answer should be 1/3(t-17)(2t+3)^3/2 – 2/25(2t+3)^5/2

  7. Madison Auten
    May 1, 2013 at 10:52 am

    For #7 a, how do you know which equation is supply and which is demand?

    • Nicholas Long
      May 1, 2013 at 3:20 pm

      The best way to tell is to graph both equations, the one that slopes downward is demand and the one that slopes upward is supply.

    • May 2, 2013 at 1:29 pm

      I always start by drawing the basic supply and demand curve and then graph the points and compare the two! Just another way to look at it.

  8. Carson Armentrout
    May 1, 2013 at 9:55 pm

    Can someone give me some help with 1 i, I just can’t seem to get it

    • Nick Thunman
      May 1, 2013 at 11:31 pm

      This one uses the (A10) rule: v(x)^n v'(x). You have to use the put and take technique he showed us in class for the v(x)^n function.

  9. Nick Thunman
    May 1, 2013 at 11:27 pm

    If anybody would like to meet tomorrow at Thomas Cooper to do some review problems for any of the sections, I was hoping to get a study room and do some work. You can text me tonight or tomorrow, my number is 703-307-6909. If tomorrow doesn’t work, Friday would be good, too.

  10. Michelle Velez- Martinez
    May 2, 2013 at 1:30 pm

    Can anyone give me the correct answer for number 10? I feel like I got it, but I would just like to verify the answer.

  11. Lauren Bennett
    May 2, 2013 at 9:28 pm

    I went to ACE Tutoring today and figured out all the integral problems for #1, Here are the answers:

    A: 27lnt(-1/3t)+3t^3
    B: 1/13ln[13x^4-7] (absolute value)
    C: 2/3(3x^2+1)^1/2
    D: 1/5e^5x^2
    E: 1/15(2t+3)^5/2
    F: (lnx)(4/3x^3)-4/9x^3
    G: 1/4ln[x^4+4x+23] (absolute value)
    H: (27t^2)(1/23e^23t)-5t(1/529)e^23t-(1/529)(1/23)e^23t This is a two step problem, you have to integrate twice.
    I: 1/48(6x^2-8x)^24

    Im pretty sure all of these are right, Hopefully these answers will help everyone!

    • Carly Gillet
      May 3, 2013 at 7:12 pm

      For G, i got 1/2 ln[x^2+4x+23]. Do you know where you got the 1/4 and the x^4 from? I was pretty sure I had the right answer.

  12. Morgan Leiter
    May 2, 2013 at 11:38 pm

    I don’t understand how you got 1/48(6x^2-8x)^24 for question 1, letter i. I thought it was 1/24 not 1/48.

    • Brooke Balentine
      May 3, 2013 at 10:32 am

      I got 1/48 too.To get 1/48 you had to put and take the (1/2) and 2 but you also had to grab the ^23 from the problem and use the rule where you add one to that and then multiply the equation by (1/n+1) so you would have (1/2)(1/24)(6x^2-8x)^24 Hope that helped and wasn’t too confusing the way I explained it.

      • Carly Gillet
        May 3, 2013 at 7:16 pm

        This is what I got too!

  13. Christina Gleaton
    May 3, 2013 at 1:10 pm

    For 1B, would it be best to try and multiply to get what you need on the top and bottom, or would it be easier to just use integration by parts?

    • Nick Calcanes
      May 3, 2013 at 1:57 pm

      You could pull out the 1/13 which would leave you just x^4-7 in the denominator and your numerator would stay the same… At that point you’d be able to do the 1/13*ln of your denominator because you’d have the perfect derivative of it on the top.

  14. karlieruth
    May 3, 2013 at 1:17 pm

    For number 1 part H, I know that it is a two step integration problem. I got the u, du, and v for the second part I just need help figuring out how to solve it?

    Karlie Bennett
    Lauren Baker

  15. Avery Ducharme
    May 3, 2013 at 2:43 pm

    does anyone know the answer for number 2???

    • Alyssa Aronson
      May 3, 2013 at 4:05 pm

      number 2 part b is F(1) = 4.5 and F(.5)= 1.75 or 7/4. so F(1)-F(.5)=2.75!

      If you were having trouble finding the antiderivative it is F(x)= t^3 + 1/2 t^2 + 3x

      • Carly Gillet
        May 3, 2013 at 9:03 pm

        Do you know the answer for part a? Did you start by factoring out (7-x^2)(7-x^2)

  16. Kellie Hafko
    May 3, 2013 at 3:33 pm

    For number 4, is that just another definite integral problem from negative square root of 2 to square root of 2?

    • Alyssa Aronson
      May 3, 2013 at 4:06 pm

      yeah you do it the same way you would do number 2.

  17. Thomas Funderburk
    May 3, 2013 at 5:23 pm

    I am unclear on the steps for 1a. I see the answer but not sure how you get it with the ln in the problem.

  18. Joey Fantasy
    May 3, 2013 at 6:03 pm

    I need help with 1.F please

  19. will grainger
    May 3, 2013 at 6:15 pm

    Hey Joe, I see you need help with 1.F.
    First this is an intergration by parts problem.
    Make lnx your u and find the derivative which is 1/x
    Next make 4x^2 your dv and integrate it as 4/3(x^3)
    plug in your values as follows : lnx 4/3(x^3) – S 4/3(x^3) dx/x
    lnx 4/3(x^3) – 4/3 S x^2 the bottom x on the dx removes a degree from the integral.
    and finally lnx 4/3(x^3) – 4/3 ((x^3)/ 3)

    let me know if this helps and good look on your exam

  20. Christopher James
    May 3, 2013 at 6:40 pm

    Does anyone know how to do number 10?

  21. Nick Thunman
    May 3, 2013 at 7:14 pm

    I typed up all of the derivative, antiderivative and integral notes, so if anybody would like me to email them to you just let me know. Also, text me if you would like to meet at the library to review tonight.

  22. joey Fantasy
    May 3, 2013 at 8:59 pm

    thanks grainger

  23. Thomas Funderburk
    May 3, 2013 at 9:54 pm

    Can someone help me with 8 and 9

  24. Bailey Burch
    May 4, 2013 at 9:42 am

    8 and 9?

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