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Math Help - A few Webwork Integral problems have me stuck.

  1. #1
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    A few Webwork Integral problems have me stuck.

    I am working on substitution and I get far with each question but then I get near the end of evaluating the integrals with differentiation (or antiderivatives, can't remember off the top off my head) and I get stuck or get the wrong answer.


    here are the ones I am stuck on(I will try multiple forms of writing it):

    1.)

    Evaluate the indefinite integral:

    Integral bar --cos(x)-- dx/1 = _______________ + C
    -----------4sin(x)+8


    (I started by substituting "4sin(x) + 8" with "u".

    I got in the end (1/4)(1/(-cos(x)+8)) + C .



    2.)

    Evaluate the definite integral


    Integral bar High# e^6, Low#1 or [1,e^6] if you prefer.----dx-----
    ------------------------------------------------------x(1+ln(x))

    = _______________

    I substituted "u" for 1+ln(x) because 1/x can be used to replace the 1/x in the main integral.

    I worked this one so much I got lost.

    u = 1 + ln(x)
    du = 1/x dx or dx/x

    = integral bar [1,7] (<= replacing the old values since I changed it.) du 1/u
    =-(1+1/x) and a couple other versions as well.



    3.) This one stumped me entirely.

    Find the following indefinite integral

    integral bar ----x----- dx
    --------------sqrt(x+8)


    The last isn't too hard to get the first part but then I tried the second part the same way and I got a wrong answer.

    4.) If f(x) = integral bar [x,x^2] (t^2)dt , then

    f '(-3)= _________.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Whitewolfblue View Post
    I am working on substitution and I get far with each question but then I get near the end of evaluating the integrals with differentiation (or antiderivatives, can't remember off the top off my head) and I get stuck or get the wrong answer.


    here are the ones I am stuck on(I will try multiple forms of writing it):

    1.)

    Evaluate the indefinite integral:

    Integral bar --cos(x)-- dx/1 = _______________ + C
    -----------4sin(x)+8


    (I started by substituting "4sin(x) + 8" with "u".

    I got in the end (1/4)(1/(-cos(x)+8)) + C .
    This is incorrect.

    \int \frac { \cos x}{4 \sin x + 8}dx

    Let u = 4 \sin x + 8

    \Rightarrow du = 4 \cos x dx

    \Rightarrow \frac {1}{4} du = \cos x dx

    So our integral becomes:

    \frac {1}{4} \int \frac {1}{u}du

    Now continue....



    2.)

    Evaluate the definite integral


    Integral bar High# e^6, Low#1 or [1,e^6] if you prefer.----dx-----
    ------------------------------------------------------x(1+ln(x))

    = _______________

    I substituted "u" for 1+ln(x) because 1/x can be used to replace the 1/x in the main integral.

    I worked this one so much I got lost.

    u = 1 + ln(x)
    du = 1/x dx or dx/x

    = integral bar [1,7] (<= replacing the old values since I changed it.) du 1/u
    =-(1+1/x) and a couple other versions as well.
    \int_{1}^{e^6} \frac {1}{x(1 + \ln x)}dx

    Let u = 1 + \ln x

    \Rightarrow du = \frac {1}{x}dx

    Now, when x=1, u = 1 + \ln 1 = 1 + 0 = 1

    when x = e^6, u = 1 + \ln \left( e^6 \right) = 7

    So our integral becomes:

    \int_{1}^{7} \frac {1}{u}du

    Now continue



    3.) This one stumped me entirely.

    Find the following indefinite integral

    integral bar ----x----- dx
    --------------sqrt(x+8)
    \int \frac {x}{ \sqrt {x + 8}}dx

    Let u = x + 8 \mbox {      } \Rightarrow x = u - 8

    \Rightarrow du = dx

    So our integral becomes:

    \int \frac {u - 8}{ \sqrt {u}}du

    = \int \frac {u - 8}{u^{ \frac {1}{2}}}du

    = \int \left( u^{ \frac {1}{2}} - 8u^{- \frac {1}{2}} \right) du

    Now continue...


    The last isn't too hard to get the first part but then I tried the second part the same way and I got a wrong answer.

    4.) If f(x) = integral bar [x,x^2] (t^2)dt , then

    f '(-3)= _________.
    We need the second fundamental theorem of calculus here, i'm not that good with it.

    f(x) = \int_{x}^{x^2} t^2 dt = \int_{x}^{0} t^2 dt + \int_{0}^{x^2} t^2 dt

    = - \int_{0}^{x} t^2 dt + \int_{0}^{x^2} t^2 dt

    Apply the second fundamental theorem of calculus:

    \Rightarrow f'(x) = \frac {d}{dt} \left( \int_{0}^{x^2} t^2 dt - \int_{0}^{x} t^2 dt \right)

    = \left( x^2 \right)^2 \cdot 2x - x^2

    = 2x^5 - x^2

    Now just find f'(-3) .......i did pretty much the whole problem, i leave this last bit to you

    I guess you could check this manually, it won't be too much work in this case to actually find the integral, evaluate between the limits and then differentiate again. good luck

    if you get stuck with any again, say so

    EDIT: I verified the last one myself, it checks out
    Last edited by Jhevon; May 29th 2007 at 02:32 PM.
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  3. #3
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    Hello, Whitewolfblue!

    Another approach to #3 . . .


    3)\;\int\frac{x}{\sqrt{x+8}}\,dx

    Let \sqrt{x+5} \,= \,u\quad\Rightarrow\quad x \,= \,u^2 - 5\quad\Rightarrow\quad dx \,= \,2u\,du

    Substitute: . \int\frac{u^2-5}{u}(2u\,du)\;=\;2\int(u^2-5)\,du \;=\;2\left(\frac{u^3}{3} - 5u\right) + C

    Factor: . \frac{2}{3}u(u^2 - 15) + C

    Back-substitute: . \frac{2}{3}\sqrt{x+5}\left([\sqrt{x+5}]^2 - 15\right) + C \;=\;\frac{2}{3}\sqrt{x+5}\,(x + 5 - 15) + C

    Answer: . \frac{2}{3}\sqrt{x+5}\,(x - 10) + C

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  4. #4
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    Question

    Thanks Soroban for that interesting way. It gave me an idea for a separate problem and I got it right doing so. Unfortunately trying it your way for that exact problem I was given and "Incorrect" message.


    I got as far as Jhevon demonstrated on questions 2,3 and 4 but I am still stuck.

    Kiba
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Whitewolfblue View Post
    Thanks Soroban for that interesting way. It gave me an idea for a separate problem and I got it right doing so. Unfortunately trying it your way for that exact problem I was given and "Incorrect" message.


    I got as far as Jhevon demonstrated on questions 2,3 and 4 but I am still stuck.

    Kiba
    if you got as far as i did, that means you didn't do anything. you should have any problem picking up the problems where i left off.

    2)
    we left off here:

    \int_{1}^{7} \frac {1}{u}du .....the integral of \frac{1}{u} is \ln u

    \Rightarrow \int_{1}^{7} \frac {1}{u}du = \left[ \ln u \right]_{1}^{7}

    = \ln 7 - \ln 1 = \ln 7


    3)
    we left off at

    \int \left( u^{ \frac {1}{2}} - 8u^{- \frac {1}{2}} \right) du ......we integrate using the power rule

    = \frac {2}{3}u^{ \frac {3}{2}} - 16u^{ \frac {1}{2}} + C

    4)
    we left off at

    f'(x) = 2x^5 - x^2

    \Rightarrow f'(-3) = 2(-3)^5 - (-3)^2

    = -486 - 9

    = -495 .......you especially shouldn't have any problems with this one
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  6. #6
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    A new one of integration by parts

    I am having trouble with another integral problem.


    Use integration by parts to evaluate the integral:

    integral bar [1,4] sqrt(t) ln(t) dt =______________



    I did it this way:

    u = ln(t) | dv = t^.5
    du = 1/x dx | v = (t^1.5)/1.5

    Using : uv - integral bar v du

    (ln(t))((t^1.5)/1.5)-(ln(t))((t^1.5)/1.5)-integral bar[1,4] (t^1.5)/1.5 (1/x) dx


    This is what I got
    ((ln(4))(4^1.5/1.5)-(ln(1))(1^1.5/1.5))-((4^2.5/2.5)(ln(4))-(1^2.5/2.5)(ln(1)))

    or -10.351


    I am thinking maybe I am supposed to use the product rule but I am unsure exactly.

    Kiba
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  7. #7
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    You mean \int_1^4\sqrt t\ln t\,dt ??

    Try to use LaTeX please.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Whitewolfblue View Post
    I am having trouble with another integral problem.


    Use integration by parts to evaluate the integral:

    integral bar [1,4] sqrt(t) ln(t) dt =______________



    I did it this way:

    u = ln(t) | dv = t^.5
    du = 1/x dx | v = (t^1.5)/1.5

    Using : uv - integral bar v du

    (ln(t))((t^1.5)/1.5)-(ln(t))((t^1.5)/1.5)-integral bar[1,4] (t^1.5)/1.5 (1/x) dx


    This is what I got
    ((ln(4))(4^1.5/1.5)-(ln(1))(1^1.5/1.5))-((4^2.5/2.5)(ln(4))-(1^2.5/2.5)(ln(1)))

    or -10.351


    I am thinking maybe I am supposed to use the product rule but I am unsure exactly.

    Kiba
    now that i can use LaTex, i find it hard to read stuff like this, so i will do it

    You should post new questions in a new thread. the formula you used is incorrect.

    \int_{1}^{4} \sqrt {t} \ln t \mbox { } dt = \frac {2}{3}t^{ \frac {3}{2}} \ln t - \frac {2}{3} \int_{1}^{4} t^{ \frac {3}{2}} \cdot \frac {1}{t} \mbox { } dt

    = \frac {2}{3}t^{ \frac {3}{2}} \ln t - \frac {2}{3} \int_{1}^{4} t^{ \frac {1}{2}} \mbox { } dt

    = \left[ \frac {2}{3}t^{ \frac {3}{2}} \ln t - \frac {4}{9}t^{ \frac {3}{2}} \right]_{1}^{4}

    = \frac {2}{3}4^{ \frac {3}{2}} \ln 4 - \frac {4}{9} 4^{ \frac {3}{2}} + \frac {4}{9}

    \approx 4.2825

    If you are unclear about anything, please ask. i don't like using the formula, so i just did it directly.

    PS. i know Krizalid told me how to align things, but i still haven't learnt the code yet, i'll have to review it
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    now that i can use LaTex, i find it hard to read stuff like this, so i will do it
    Did you know that MathType works as a LaTeX translator?
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  10. #10
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    I apologize I am new to Mathtype I was wondering how everyone was able to create pictures.

    I will try and use IT next time.

    When I tried it a seperate way I got a negative version of what you got.

    thanks again

    Kiba
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Did you know that MathType works as a LaTeX translator?
    i didn't know that, how do you access that feature?
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Whitewolfblue View Post
    I apologize I am new to Mathtype I was wondering how everyone was able to create pictures.

    I will try and use IT next time.

    When I tried it a seperate way I got a negative version of what you got.

    thanks again

    Kiba
    chances are you used the wrong formula again. look up the formula and make sure you're using it properly. on the interval [1,4] the graph \sqrt {t} \ln t is above the x-axis, any negative answer is wrong

    the formula is:

    \int u \cdot dv = uv - \int v \cdot du

    that is the version i see used most often

    i like this version better

    \int uv' = uv - \int u'v
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  13. #13
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    OK

    I assume you have full MathType version.

    .- Open it
    .- Preferences
    .- Translators...
    .- Select "translation to other language (text)"
    .- Select the translator TeX--AMS-LaTeX
    .- Deactivate "include MathType data in translation"

    Then, for example, write something, and you paste it into the forum

    \[
    \int {(x + 1)dx}
    \]

    Remove the \[ and \], and enclose the rest on math /math; finally, you'll have \int {(x + 1)dx}
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    OK

    I assume you have full MathType version.

    .- Open it
    .- Preferences
    .- Translators...
    .- Select "translation to other language (text)"
    .- Select the translator TeX--AMS-LaTeX
    .- Deactivate "include MathType data in translation"

    Then, for example, write something, and you paste it into the forum

    \[
    \int {(x + 1)dx}
    \]

    Remove the \[ and \], and enclose the rest on math /math; finally, you'll have \int {(x + 1)dx}
    ah! I see! thanks!
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