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Math Help - Integral

  1. #1
    Member srirahulan's Avatar
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    Exclamation Integral

    using integration by parts find out the value of this ,i go by this way then i stucked>>>>>>
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    Re: Integral

    Quote Originally Posted by srirahulan View Post
    using integration by parts find out the value of this ,
    Why use parts? What is the derivative of $\dfrac{2^{x+1}}{\ln(2)}~?$
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    Re: Integral

    [QUOTE=srirahulan;810859]using integration by parts find out the value of this ,[/QUOTE

    Please learn to use the new TeX \$\dfrac{2^{x+1}}{\ln(2)}\$

    OR \$\$\dfrac{2^{x+1}}{\ln(2)}\$\$ gives $$\dfrac{2^{x+1}}{\ln(2)}$$
    Last edited by Plato; February 20th 2014 at 06:08 AM.
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  4. #4
    Member srirahulan's Avatar
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    Lightbulb Re: Integral

    Quote Originally Posted by srirahulan View Post
    using integration by parts find out the value of this ,i go by this way then i stucked>>>>>>
    Here i forgot to put sqrt in there>>>>
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    Re: Integral

    Quote Originally Posted by srirahulan View Post
    Here i forgot to put sqrt in there>>>>
    You did not answer the question I posted. Why?
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  6. #6
    Member srirahulan's Avatar
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    Re: Integral

    So How can i do that easily??? ((please note i have correction in my problem))
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  7. #7
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    Re: Integral

    Quote Originally Posted by srirahulan View Post
    So How can i do that easily??? ((please note i have correction in my problem))
    Please learn to use the new TeX \$\dfrac{2^{x+1}}{\ln(2)}\$

    OR \$\$\dfrac{2^{x+1}}{\ln(2)}\$\$ gives $$\dfrac{2^{x+1}}{\ln(2)}$$

    Please, please take a look at this. Also look at this.

    Your answer is $\displaystyle\left. {{{{2^{x + 1}}} \over {\ln (2)}}} \right|_{ - 1}^1$


    There is no need for parts in this question.
    Last edited by Plato; February 20th 2014 at 03:28 PM.
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  8. #8
    Member srirahulan's Avatar
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    Arrow Re: Integral

    Thank You Plato>>

    I have problem like this,
    First they ask me to proof this $$\frac{d\frac{2^{x}}{ln2}}{dx}=2^{x}$$,Then they ask to find out $$\int2^{x}$$,After that Using the parts$$\int_{-1}^{1}2^{\sqrt{2x+1}}$$

    In this problem i must use parts>>>>
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  9. #9
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    Re: Integral

    Quote Originally Posted by srirahulan View Post
    Thank You Plato>>

    I have problem like this,
    First they ask me to proof this $$\frac{d\frac{2^{x}}{ln2}}{dx}=2^{x}$$,Then they ask to find out $$\int2^{x}$$,After that Using the parts$$\int_{-1}^{1}2^{\sqrt{2x+1}}$$

    In this problem i must use parts>>>>
    Look at this.
    Last edited by Plato; February 20th 2014 at 03:42 PM.
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  10. #10
    Member srirahulan's Avatar
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    Re: Integral

    Explain How you use parts?? please explain
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    Re: Integral

    Quote Originally Posted by srirahulan View Post
    Explain How you use parts?? please explain
    Integral-wolframalpha-integral_2sqrt2x1_dx-2014-02-20_1923.gif
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  12. #12
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    Re: Integral

    [
    Quote Originally Posted by srirahulan View Post
    Thank You Plato>>

    I have problem like this,
    First they ask me to proof this $$\frac{d\frac{2^{x}}{ln2}}{dx}=2^{x}$$,Then they ask to find out $$\int2^{x}$$
    Okay, knowing that the derivative of \frac{2^x}{ln 2} is 2^x, it follows immediately that \int 2^x dx= \frac{2^x}{ln(2)}+ C

    ,After that Using the parts$$\int_{-1}^{1}2^{\sqrt{2x+1}}$$

    In this problem i must use parts>>>>
    I would first use a substitution. Since I now know how to integrate [latex]2^x[/latex] I would get back to that by letting y= \sqrt{2x+ 1}= (2x+ 1)^{1/2}. Then [tex]dy= (1/2)(2x+ 1)^{-1/2}(2)dx= \frac{dx}{\sqrt{2x+ 1}= \frac{dx}{y}[/latex] or [latex]ydy= dx[/latex]

    So in term of the new variable y= \sqrt{2x+ 1} the integral is \int y2^y dy and you can use "integration" by parts, taking [latex]u= y[/latex] and [latex]dv= 2^y[/latex]
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