Integral help

  • Thread starter madahmad1
  • Start date
42
0
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.
 

rock.freak667

Homework Helper
6,230
31
[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:
42
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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?
 
1,750
1
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.
 
1,750
1
[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:

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