4.2.5 AP Calculus Exam int of e

[karush]

View attachment 9321

calculator returned this but know sure why
$\displaystyle2 \int _1^2e^udu$
note there might be a duplicat of this post ?

 

[skeeter]

$u=\sqrt{x} \implies du = \dfrac{1}{2\sqrt{x}} \, dx$

\(\displaystyle {\color{red}{2}} \int_1^4 \frac{e^{\sqrt{x}}}{{\color{red}{2}} \sqrt{x}} \, dx = 2 \int_1^2 e^u \, du\)

 

[karush]

$u=\sqrt{x} \implies du = \dfrac{1}{2\sqrt{x}} \, dx$

\(\displaystyle {\color{red}{2}} \int_1^4 \frac{e^{\sqrt{x}}}{{\color{red}{2}} \sqrt{x}} \, dx = 2 \int_1^2 e^u \, du\)

Why would that change the limits,?

 

[topsquark]

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

 

[HallsofIvy]

When you change from x to u, every reference to "x" has to change to a reference to "u". "[tex]\int_1^4 dx[/tex]" means we are taking the integral fron x= 1 to x= 4. We have to change that to u. When x= 1, [tex]u= \sqrt{1}= 1[/tex] and when x= 4, [tex]u= \sqrt{4}= 2[/tex].