4.2.236 AP calculus Exam integral with u substitution

[karush]

AP Calculas Exam Problem$\textsf{Using
$\displaystyle u=\sqrt{x}, \quad
\int_1^4\dfrac{e^{\sqrt{x}}}{\sqrt{x}}\, dx$
is equal to which of the following}$
$$
(A)2\int_1^{16} e^u \, du\quad
(B)2\int_1^{4} e^u \, du\quad
(C) 2\int_1^{2} e^u \, du\quad
(D) \dfrac{1}{2}\int_1^{2} e^u \, du\quad
(E) \int_1^{4} e^u \, du\
$$
By observation
$$\int_1^4\dfrac{e^u}{u}\, du$$ok I got this far but...

 

[skeeter]

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

$\displaystyle 2\int_1^4 e^{\sqrt{x}} \cdot \dfrac{1}{2\sqrt{x}} \, dx = 2\int_1^2 e^u \, du$

 

[Greg]

\(\displaystyle u=e^{\sqrt x}\)

\(\displaystyle 2\,du=\frac{e^{\sqrt x}}{\sqrt x}\,dx\)

\(\displaystyle 2\int_e^{e^2}\,du=2e(e-1)\)

 

[HallsofIvy]

In addition to "observation" you need to do a little Calculus. You have replaced "[tex]\sqrt{x}[tex]" with "$$u$$" but with $$u= \sqrt{x}$$, du is not dx!

 

[karush]

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

$\displaystyle 2\int_1^4 e^{\sqrt{x}} \cdot \dfrac{1}{2\sqrt{x}} \, dx = 2\int_1^2 e^u \, du$

I think this was the best way to do it.