Integrals #6 and #7

May 2009
959
362
Challenge Problems:


\(\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx \)


\(\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx \)



EDIT: I don't like the second integral. It doesn't even make any sense unless you state that \(\displaystyle \sqrt[3]{2x^{3}-3x^2-x+1} \) is a real-valued function. Sorry. It's a bad problem.



Moderator edit: Approved Challenge question.
 
Last edited by a moderator:

chiph588@

MHF Hall of Honor
Sep 2008
1,163
429
Champaign, Illinois
Challenge Problems:


\(\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx \)
\(\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx =0 \)

A simple Taylor/Laurent expansion of \(\displaystyle e^{x /a} \) and \(\displaystyle e^{b /x} \) should prove my assertion.
 
May 2009
959
362
\(\displaystyle \int^{b}_{a} \frac{e^{x /a} - e^{b /x}}{x} \ dx =0 \)

A simple Taylor/Laurent expansion of \(\displaystyle e^{x /a} \) and \(\displaystyle e^{b /x} \) should prove my assertion.
Could you elaborate?

The solution I had in mind is even simpler than that.
 

simplependulum

MHF Hall of Honor
Jan 2009
715
427
If my calculations are correct , the integrals are both zero .


For the first one ,

Sub. \(\displaystyle x = \frac{ab}{t} \) in the second integral .


and we will find that it is equal to

\(\displaystyle \int_a^b \frac{e^{x/a}}{x}~dx - \int_a^b \frac{e^{t/a}}{t}~dt \)

And the second integral , i find that it is identical to

\(\displaystyle \sqrt[3]{(2x-1)[(x-\frac{1}{2})^2 - \frac{5}{4}]} \)

Sub. \(\displaystyle x = 1-t \) and we have

\(\displaystyle I = -I \)

\(\displaystyle I = 0 \)
 
  • Like
Reactions: chiph588@
May 2009
959
362
If my calculations are correct , the integrals are both zero .


For the first one ,

Sub. \(\displaystyle x = \frac{ab}{t} \) in the second integral .


and we will find that it is equal to

\(\displaystyle \int_a^b \frac{e^{x/a}}{x}~dx - \int_a^b \frac{e^{t/a}}{t}~dt \)

And the second integral , i find that it is identical to

\(\displaystyle \sqrt[3]{(2x-1)[(x-\frac{1}{2})^2 - \frac{5}{4}]} \)

Sub. \(\displaystyle x = 1-t \) and we have

\(\displaystyle I = -I \)

\(\displaystyle I = 0 \)

Nice.

If for the first integral you made the same substitution but didn't break it up into two integrals, you would get I = -I.

And for the second integral, Maple claims the answer is a complex number and Mathematica doesn't know what to do with it. The issue is the cube root of negative real numbers.
 

chiph588@

MHF Hall of Honor
Sep 2008
1,163
429
Champaign, Illinois
\(\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx \)
Let \(\displaystyle x=t+\tfrac12 \)

We get \(\displaystyle \int^{1}_{0} \sqrt[3]{2x^{3}-3x^2-x+1} \ dx = \int^{\tfrac12}_{-\tfrac12} \sqrt[3]{2(t+\tfrac12)^{3}-3(t+\tfrac12)^2-(t+\tfrac12)+1} \ dx = \) \(\displaystyle \int^{\tfrac12}_{-\tfrac12} \sqrt[3]{2t^3-\tfrac52t} \ dx = 0 \) since \(\displaystyle \sqrt[3]{2t^3-\tfrac52t} \) is odd.
 
Last edited: