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Math Help - Integration problems!

  1. #1
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    Integration problems!

    Find the continuous function f(x) and constant number a which satisfies \int_0^x {f(u)\,du} = e^x - ae^{2x} \int_0^1 {f(u) \cdot e^{ - u} \,du}

    Evaluate \int_0^\infty {\frac{x}<br />
{{e^x - 1}}\,dx}

    Evaluate \int_0^1 {\frac{{x - 1}}<br />
{{\ln x}}\,dx}

    Evaluate \int_0^\pi {xe^x \sin x\,dx}

    Evaluate \int_0^\infty {\frac{{e^{ - x} - 2e^{ - 3x} + e^{ - 5x} }}<br />
{{x^2 }}\,dx}

    Evaluate \int_0^1 {\int_0^1 {\frac{{x^3 y^3 }}<br />
{{1 - xy}}\,dx} \,dy}

    Evaluate \int_0^1 {\frac{{\arcsin x}}<br />
{x}\,dx}

    Prove that 2\int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} = \int_{ - 1}^1 {\frac{1}<br />
{{\sqrt {1 - x^2 } }}\,dx}

    Evaluate \sum\limits_{k = 0}^\infty {\left( {\frac{1}<br />
{{3k + 1}} - \frac{1}<br />
{{3k + 2}}} \right)}


    Getting rid of them I'm done! Thanks!
    Last edited by Mess; December 22nd 2007 at 10:00 AM. Reason: Adding more problems
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  2. #2
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    Quote Originally Posted by Mess View Post
    Find the continuous function f(x) and constant number a which satisfies \int_0^x {f(u)\,du} = e^x - ae^{2x} \int_0^1 {f(u) \cdot e^{ - u} \,du}
    Differentiate this with respect to x, to get:

     <br />
f(x)=e^x - 2a e^{2x} \int_0^1 f(u) e^{-u} du<br />

    Now put K=\int_0^1 f(u) e^{-u} du, then:

     <br />
f(x)=e^x - 2a K e^{2x}<br />

    which you substitute back into the definition of K to finish.

    RonL
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  3. #3
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    Slightly different from CaptainBlack's solution.

    Quote Originally Posted by Mess View Post
    Find the continuous function f(x) and constant number a which satisfies \int_0^x {f(u)\,du} = e^x - ae^{2x} \int_0^1 {f(u) \cdot e^{ - u} \,du}
    Set x=0,

    0 = 1 - a\int_0^1 {f(u) \cdot e^{ - u} \,du} \,\therefore \,a\int_0^1 {f(u) \cdot e^{ - u} \,du} = 1.

    Hence \int_0^x {f(u)\,du} = e^x - e^{2x} \implies f(x) = e^x - 2e^{2x} .

    Now this is routine, it remains to compute \int_0^1 {\left( {e^u - 2e^{2u} } \right)e^{ - u} \,du} = 3 - 2e.

    The rest follows.
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  4. #4
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    Quote Originally Posted by Mess View Post
    Evaluate \int_0^\infty {\frac{x}<br />
{{e^x - 1}}\,dx}
    Effortlessly, start by proving that (for k>0) \frac{1}<br />
{{k^2 }} = \int_0^\infty {xe^{ - kx} \,dx} .

    And from the geometric series

    \frac{1}<br />
{{e^x - 1}} = e^{ - x} \cdot \frac{1}<br />
{{1 - e^{ - x} }} = e^{ - x} \sum\limits_{n = 0}^\infty {e^{ - nx} } = \sum\limits_{k = 1}^\infty {e^{ - kx} } .

    Interchange the integral by the sum and we'll get

    \int_0^\infty {\frac{x}<br />
{{e^x - 1}}\,dx} = \int_0^\infty {x\sum\limits_{k = 1}^\infty {e^{ - kx} } \,dx} = \sum\limits_{k = 1}^\infty {\int_0^\infty {xe^{ - kx} \,dx} } = \sum\limits_{k = 1}^\infty {\frac{1}<br />
{{k^2 }}} = \frac{{\pi ^2 }}<br />
{6}.
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  5. #5
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    This kind of integral is well known. Actually, the only trick we need to apply, it's to construct a double integral. (Of course, I'd never create a double integral without switch the integration order.)

    Quote Originally Posted by Mess View Post
    Evaluate \int_0^1 {\frac{{x - 1}}<br />
{{\ln x}}\,dx}
    \frac{{x - 1}}<br />
{{\ln x}} = \int_0^1 {x^u \,du} . So,

    \int_0^1 {\frac{{x - 1}}<br />
{{\ln x}}\,dx} = \int_0^1 {\int_0^1 {x^u \,du} \,dx} = \int_0^1 {\underbrace {\int_0^1 {x^u \,dx} }_{\dfrac{1}<br />
{{u + 1}}}\,du} = \ln 2.
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  6. #6
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    Quote Originally Posted by Mess View Post
    Evaluate \int_0^1 {\int_0^1 {\frac{{x^3 y^3 }}<br />
{{1 - xy}}\,dx} \,dy}
    \frac{1}<br />
{{1 - xy}} = \sum\limits_{k = 0}^\infty {(xy)^k } . So,

    \int_0^1 {\int_0^1 {\frac{{x^3 y^3 }}<br />
{{1 - xy}}\,dx} \,dy} = \int_0^1 {\int_0^1 {\sum\limits_{k = 0}^\infty {x^{k + 3} y^{k + 3} } \,dx} \,dy} = \sum\limits_{k = 0}^\infty {\int_0^1 {\int_0^1 {x^{k + 3} y^{k + 3} \,dx} \,dy} } .

    And finally (evaluation of double integral is easy)

    \sum\limits_{k = 0}^\infty {\frac{1}<br />
{{(k + 4)^2 }}} = \sum\limits_{n = 4}^\infty {\frac{1}<br />
{{n^2 }}} = \frac{{\pi ^2 }}<br />
{6} - 1 - \frac{1}<br />
{4} - \frac{1}<br />
{9} = \frac{{\pi ^2 }}<br />
{6} - \frac{{49}}<br />
{{36}}.
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  7. #7
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    There's something to be said about plain, old, dull, numerical methods.

    \int_{0}^{1}\frac{sin^{1}(x)}{x}dx

    By the midpoint method, we get \frac{{\pi}ln(2)}{2}
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  8. #8
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    Quote Originally Posted by Mess View Post
    Evaluate \int_0^1 {\frac{{\arcsin x}}<br />
{x}\,dx}
    Well, this one is not so hard to solve.

    \arcsin x = \int_0^x {\frac{1}<br />
{{\sqrt {1 - u^2 } }}\,du} . So,

    \int_0^1 {\frac{{\arcsin x}}<br />
{x}\,dx} = \int_0^1 {\int_0^x {\frac{1}<br />
{{x\sqrt {1 - u^2 } }}\,du} \,dx} = \int_0^1 {\underbrace {\int_u^1 {\frac{1}<br />
{x}\,dx} }_{ - \ln u}\frac{1}<br />
{{\sqrt {1 - u^2 } }}\,du} , hence

    \int_0^1 {\frac{{\arcsin x}}<br />
{x}\,dx} = - \int_0^1 {\frac{{\ln u}}<br />
{{\sqrt {1 - u^2 } }}\,du} .

    Now substitute u=\sin\alpha, the integral becomes

    - \int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } = \frac{\pi }<br />
{2}\ln 2,

    which confirms galactus' answer.

    You can check the last result here. (See post #5.)
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  9. #9
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    Hey Krizmeister.

    I mostly go for the easy out, but I tried working the arcsine one out using your methods, only a slight variation. I hammered it into:

    \int_{0}^{1}\frac{1}{\sqrt{1-u^{2}}}du\cdot\int_{1}^{2}\frac{1}{x}dx=\frac{ln(2  )\pi}{2}.
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  10. #10
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    Quote Originally Posted by Mess View Post
    Prove that 2\int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} = \int_{ - 1}^1 {\frac{1}<br />
{{\sqrt {1 - x^2 } }}\,dx}
    \pi=\pi is a boring way to claim equality, so let's apply some of double integration.

    2\int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} = 2\int_0^1 {2\sqrt {1 - x^2 } \,dx} .

    Insert 2\sqrt {1 - x^2 } = \int_{x^2 }^1 {\frac{1}<br />
{{\sqrt {1 - u} }}\,du} and switch integration order

    \int_0^1 {\int_{x^2 }^1 {\frac{1}<br />
{{\sqrt {1 - u} }}\,du} \,dx} = \int_0^1 {\underbrace {\int_0^{\sqrt u } {dx} }_{\sqrt u }\frac{1}<br />
{{\sqrt {1 - u} }}\,du} .

    Now substitute u=\varphi^2, then the integral becomes

    2\int_0^1 {\frac{{\varphi ^2 }}<br />
{{\sqrt {1 - \varphi ^2 } }}\,d\varphi } = \int_{ - 1}^1 {\frac{1}<br />
{{\sqrt {1 - \varphi ^2 } }}\,d\varphi } - \int_{ - 1}^1 {\sqrt {1 - \varphi ^2 } \,d\varphi } .

    From here \int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} = \int_{ - 1}^1 {\frac{1}<br />
{{\sqrt {1 - x^2 } }}\,dx} - \int_{ - 1}^1 {\sqrt {1 - x^2 } \,dx} ,

    rearrange and we're done.
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  11. #11
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    Evaluate \int_0^\pi {xe^x \sin x\,dx}
    Good, old-fashioned parts is one way to go on this one. Though, you'll have to apply it several times.

    Let u=xe^{x}, \;\ dv=sin(x)dx, \;\ du=(x+1)e^{x}, \;\ v=-cos(x)

    -xe^{x}cos(x)+\int{e^{x}cos(x)}dx+\int{xe^{x}cos(x)  }dx

    Do the parts thing again:

    \int{xe^{x}sin(x)}dx=-xe^{x}cos(x)+e^{x}cos(x)+\int{e^{x}sin(x)}dx+\int{  xe^{x}cos(x)}dx

    Continue in this fashion and you should have a -\int{xe^{x}sin(x)}dx on the right side. Add it to both sides and divide by 2 and you should end up with:

    \frac{xe^{x}sin(x)}{2}+\frac{e^{x}cos(x)}{2}-\frac{xe^{x}cos(x)}{2}
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  12. #12
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    Quote Originally Posted by Mess View Post
    Evaluate \sum\limits_{k = 0}^\infty {\left( {\frac{1}<br />
{{3k + 1}} - \frac{1}<br />
{{3k + 2}}} \right)}
    \frac{1}<br />
{{3k + 1}} - \frac{1}<br />
{{3k + 2}} = \int_0^1 {\left( {x^{3k} - x^{3k + 1} } \right)\,dx} = \int_0^1 {x^{3k} \left( {1 - x} \right)\,dx}

    Interchange sum & integral

    \sum\limits_{k = 0}^\infty {\left( {\frac{1}<br />
{{3k + 1}} - \frac{1}<br />
{{3k + 2}}} \right)} = \sum\limits_{k = 0}^\infty {\left( {\int_0^1 {x^{3k} \left( {1 - x} \right)\,dx} } \right)} = \int_0^1 {\sum\limits_{k = 0}^\infty {\Big[x^{3k} \left( {1 - x} \right)\Big]\,dx} } .

    Hence \sum\limits_{k = 0}^\infty {\left( {\frac{1}<br />
{{3k + 1}} - \frac{1}<br />
{{3k + 2}}} \right)} = \int_0^1 {\frac{{1 - x}}<br />
{{1 - x^3 }}\,dx} = \int_0^1 {\frac{1}<br />
{{1 + x + x^2 }}\,dx} .

    The rest follows.
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  13. #13
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    Quote Originally Posted by Mess View Post
    Evaluate \int_0^\infty {\frac{{e^{ - x} - 2e^{ - 3x} + e^{ - 5x} }}<br />
{{x^2 }}\,dx}
    See post #4 (the parameter), it's all you need to solve the problem. (The interchange of integration order is justified by Tonelli's Theorem, since the function is positive.)
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  14. #14
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    - \int_0^{\pi /2} {\ln (\sin \alpha )\,d\alpha } = \frac{\pi }<br />
{2}\ln 2,
    I wanted to show my method of doing this integral. It's been bugging me. No, I am not vying for Best Integrator. I am sure Big K has that in the bag. I, like K, just enjoy doing these integrals. They are like little puzzles to figure out. I see why Plato says it is rather pointless, but why not just for fun and not practicality?.

    \int_{0}^{\frac{\pi}{2}}ln(sin({\theta}))d{\theta}

    First, I use the identity: sin{\theta}=2sin(\frac{\theta}{2})cos(\frac{\theta  }{2})

    I=\int_{0}^{\frac{\pi}{2}}ln(sin{\theta})d{\theta}  =\int_{0}^{\frac{\pi}{2}}ln(2sin\frac{\theta}{2}co  s\frac{\theta}{2})d{\theta}

    =\int_{0}^{\frac{\pi}{2}}ln2d{\theta}+\int_{0}^{\f  rac{\pi}{2}}ln(sin\frac{\theta}{2})d{\theta}+\int_  {0}^{\frac{\pi}{2}}ln(cos\frac{\theta}{2})d{\theta  }

    Now, make the sub u=\frac{\theta}{2}, and we get:

    I=\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{4}}ln(s  in(u))du+2\int_{0}^{\frac{\pi}{4}}ln(cos(u))du

    Now, use the identity cos(u)=sin(\frac{\pi}{2}-u)

    and z=\frac{\pi}{2}-u.

    We get:

    \int_{0}^{\frac{\pi}{4}}ln(cos(u))du = -\int_{\frac{\pi}{2}}^{\frac{\pi}{4}}ln(sin(z))dz = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(sin(z))dz

    So, we get:

    I=\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{4}}ln(s  in(u))du+2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}ln(s  in(u))du

    =\frac{\pi}{2}ln(2)+2\int_{0}^{\frac{\pi}{2}}ln(si  n({\theta}))d{\theta}=\frac{\pi}{2}ln(2)+2I

    Solve for I and get:

    \boxed{I = -\frac{\pi}{2}ln(2)}

    Of course, the negative at the beginning results in a positive solution.
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