Expert: Paul Klarreich - 12/7/2009

Question
QUESTION: Hello, In class we have gone over integration, with "u" substitution in definite and indefinite integrals. But I can't seem to get this on.
Find the area of
y= e^(sqrt x) (from 0 to 1)
Sorry But I don't really know what to do with that problem.
Thank you.

ANSWER: Questioner: Jacob
Country: United States
Category: Calculus
Private: No
Subject: Calculus I - Definite integrals.
Question: Hello, In class we have gone over integration, with "u" substitution in definite and indefinite integrals. But I can't seem to get this on.
Find the area of
y= e^(sqrt x) (from 0 to 1)
Sorry But I don't really know what to do with that problem.
Thank you.
...............................................

Let t = sqrt(x)

t^2 = x

2t dt = dx

{
| e^(sqrt(x)) dx =
}

{
| 2 e^t t dt
}

Now do integration by parts.  Use  u = t,  dv = e^t dt.

Hint: Try THE INTEGRATOR at

integrals.wolfram.com/index.jsp

for quick and dirty answers, which might provide hints about what to do.


---------- FOLLOW-UP ----------

QUESTION: The problem is that, we haven't learned integration by parts. So we can't use it. Is that the only way to solve it?

Answer
Maybe you can try:

Let   u = sqrt(x)

du = 1/(2 sqrt(x)) dx

dx = 2 sqrt(x) du = 2 u du

{
| e^(sqrt(x)) dx =
}

{
| e^u 2u du =  Oops, that's the same as we did before.
}
........................
OK, try something else.

Let  u = e^sqrt(x)

du = e^sqrt(x)/(2 sqrt(x))  dx

     2 sqrt(x) du
dx = --------------
       e^sqrt(x)  


     2 sqrt(x) du
dx = --------------
         u

{   2 sqrt(x) du
| u -----------  =
}        u  


{  
|   2 sqrt(x) du
}        

Now  if  e^sqrt(x) = u, sqrt(x) = ln u, so you have:


{  
|   2 ln u du
}        

Aha!  Now all you have to do is integrate  ln u.  

Oh, shoot! That requires I.B.P., too.