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Math Help - Integrals

  1. #1
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    Integrals

    Hi,

    Q1. Evaluate the following integrals by suing the given substitution.

    a) f [ {e^squareroot(x)}/{2*squareroot(x)} ], u =squareroot(x)

    b) f { x*sin(x^2) } dx, substitution u = x^2

    c) f (2x+5)/(x^2 + 5x + 17), substitution u = x^2 + 5x +17

    (*note: the functions above are antiderivatives...)


    can anyone help me with these questions? im clueless - steps and answers would be appreciated...

    thanks,

    moon
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  2. #2
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    Quote Originally Posted by Moon Hoplite View Post
    Hi,

    Q1. Evaluate the following integrals by suing the given substitution.

    a) f [ {e^squareroot(x)}/{2*squareroot(x)} ], u =squareroot(x)

    b) f { x*sin(x^2) } dx, substitution u = x^2

    c) f (2x+5)/(x^2 + 5x + 17), substitution u = x^2 + 5x +17

    (*note: the functions above are antiderivatives...)


    can anyone help me with these questions? im clueless - steps and answers would be appreciated...

    thanks,

    moon
    a) \int{\frac{e^{\sqrt{x}}}{2\sqrt{x}}\,dx}.

    Notice that this can be rewritten as \int{e^{\sqrt{x}}\frac{1}{2\sqrt{x}}\,dx}.

    Like you were told, make the substitution u = \sqrt{x} = x^{\frac{1}{2}}.

    \frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}.


    So the integral reduces to

    \int{e^u\,\frac{du}{dx}\,dx}

     = \int{e^u\,du}

     = e^u + C

     = e^{\sqrt{x}} + C.
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  3. #3
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    Quote Originally Posted by Moon Hoplite View Post
    Hi,

    Q1. Evaluate the following integrals by suing the given substitution.

    a) f [ {e^squareroot(x)}/{2*squareroot(x)} ], u =squareroot(x)

    b) f { x*sin(x^2) } dx, substitution u = x^2

    c) f (2x+5)/(x^2 + 5x + 17), substitution u = x^2 + 5x +17

    (*note: the functions above are antiderivatives...)


    can anyone help me with these questions? im clueless - steps and answers would be appreciated...

    thanks,

    moon
    \int{x\sin{x^2}\,dx}.

    Use the substitution u = x^2.

    \frac{du}{dx} = 2x.


    We need to rearrange the integral so it has 2x as a factor. Notice that x = \frac{1}{2}\times 2x

    So we can rewrite this as

    \frac{1}{2}\int{2x\sin{x^2}\,dx}

     = \frac{1}{2}\int{\sin{u}\,\frac{du}{dx}\,dx}

     = \frac{1}{2}\int{\sin{u}\,du}

     = -\frac{1}{2}\cos{u} + C

     = -\frac{1}{2}\cos{x^2} + C.
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  4. #4
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    Quote Originally Posted by Moon Hoplite View Post
    Hi,

    Q1. Evaluate the following integrals by suing the given substitution.

    a) f [ {e^squareroot(x)}/{2*squareroot(x)} ], u =squareroot(x)

    b) f { x*sin(x^2) } dx, substitution u = x^2

    c) f (2x+5)/(x^2 + 5x + 17), substitution u = x^2 + 5x +17

    (*note: the functions above are antiderivatives...)


    can anyone help me with these questions? im clueless - steps and answers would be appreciated...

    thanks,

    moon
    c) \int{\frac{2x + 5}{x^2 + 5x + 17}\,dx}.

    First, rewrite this as

    \int{\frac{1}{x^2 + 5x + 17}(2x + 5)\,dx}.

    Make the substitution u = x^2 + 5x + 17 so that \frac{du}{dx} = 2x + 5.

    So the integral becomes

    \int{\frac{1}{u}\,\frac{du}{dx}\,dx}

     = \int{\frac{1}{u}\,du}

     = \ln{|u|} + C

     = \ln{|x^2 + 5x + 17|} + C.
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