Evaluating Integral: ∫ (e√x / √x) dx

[madahmad1]

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]

\int \frac{e^{\sqrt{x}}}{\sqrt{x}} dx

Let t=\sqrt{x} \Rightarrow \frac{dt}{dx}=\frac{1}{2\sqrt{x}} \Rightarrow 2 dt=\frac{dx}{\sqrt{x}}

Can you take it from here?

 

[madahmad1]

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?

 

[rocomath]

No ... if you u-sub the sqrt of x in the numerator and take the derivative, a part of your integral will appear which let's you get rid of the one in the denominator.

 

[rocomath]

\int e^{\sqrt{x}}\frac{dx}{\sqrt x}

u=\sqrt x
du=\frac{dx}{2\sqrt x} \rightarrow 2du=\frac{dx}{\sqrt x}

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

2\int e^u du