
MHB Master
#1
October 25th, 2019,
21:32
calculator returned this but know sure why
$\displaystyle2 \int _1^2e^udu$
note there might be a duplicat of this post ???

October 25th, 2019 21:32
# ADS
Circuit advertisement

MHB Journeyman
#2
October 25th, 2019,
22:45
$u=\sqrt{x} \implies du = \dfrac{1}{2\sqrt{x}} \, dx$
$ \displaystyle {\color{red}{2}} \int_1^4 \frac{e^{\sqrt{x}}}{{\color{red}{2}} \sqrt{x}} \, dx = 2 \int_1^2 e^u \, du$

MHB Master
#3
October 26th, 2019,
03:21
Thread Author
Originally Posted by
skeeter
$u=\sqrt{x} \implies du = \dfrac{1}{2\sqrt{x}} \, dx$
$ \displaystyle {\color{red}{2}} \int_1^4 \frac{e^{\sqrt{x}}}{{\color{red}{2}} \sqrt{x}} \, dx = 2 \int_1^2 e^u \, du$
Why would that change the limits,?

MHB Master
#4
October 26th, 2019,
06:58
Originally Posted by
karush
Why would that change the limits,?
The original integral over x was done over an interval (1, 4).
When we changed the variable to u(x) we are no longer integrating over x. We are now integrating over u. So the interval changes to (1, 2).
Dan

#5
October 26th, 2019,
07:53
When you change from x to u, every reference to "x" has to change to a reference to "u". "$ \displaystyle \int_1^4 dx$" means we are taking the integral fron x= 1 to x= 4. We have to change that to u. When x= 1, $ \displaystyle u= \sqrt{1}= 1$ and when x= 4, $ \displaystyle u= \sqrt{4}= 2$.