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Integral help

  1. Mar 9, 2008 #1
    If you are told to evaluate the integral and are given this problem:

    ∫ (e√x / √x) dx

    This reads that the integral of (e) to the power of square root of x divided by the square root of x multiplied by the derivative of x is……

    I have tried solving it and came up with this, is this correct if not please show me what I did wrong and how I can solve it.

    y = x2 + 2

    y = sin x

    0 ≤ x ≤ π
    π π π π
    A(D) = ∫0 (x2 + 2 – sin x) dx = ∫0 x2dx + 2 ∫0 dx - ∫0 sin x dx =

    π π π
    [x 3 / 3 ]0 + 2[x]0 + [cos x]0 = π/3 + 2π – 2

    π
    [cos x]0 = cos π – cos 0 = -1 – 1 = -2

    Is this correct?

    the integral goes from zero to pi. and x2 is x squared. Any help is appreciated.
     
  2. jcsd
  3. Mar 9, 2008 #2

    rock.freak667

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    [tex]\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx[/tex]

    Let [itex]t=\sqrt{x} \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}} \Rightarrow 2 dt=\frac{dx}{\sqrt{x}}[/itex]

    Can you take it from here?
     
    Last edited: Mar 9, 2008
  4. Mar 12, 2008 #3
    now do you turn the square root of x into the power of 1/2, then you raise it to the numerator it becomes -1/2, the direvative of that is 1/2 divided by one whic is one half. correct?
     
  5. Mar 12, 2008 #4
    No ... if you u-sub the sqrt of x in the numerator and take the derivative, a part of your integral will appear which lets you get rid of the one in the denominator.
     
  6. Mar 12, 2008 #5
    [tex]\int e^{\sqrt{x}}\frac{dx}{\sqrt x}[/tex]

    [tex]u=\sqrt x[/tex]
    [tex]du=\frac{dx}{2\sqrt x} \rightarrow 2du=\frac{dx}{\sqrt x}[/tex]

    Now make your substitutions ... hope this is a little more clear.

    [tex]2\int e^u du[/tex]
     
    Last edited: Mar 12, 2008
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