MHB 4.2.236 AP calculus Exam integral with u substitution

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

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

$$2\,du=\frac{e^{\sqrt x}}{\sqrt x}\,dx$$

$$2\int_e^{e^2}\,du=2e(e-1)$$
 
In addition to "observation" you need to do a little Calculus. You have replaced "\sqrt{x}" with "u" but with u= \sqrt{x}, du is <b>not</b> dx!
 
skeeter said:
$u = \sqrt{x} \implies du = \dfrac{1}{2\sqrt{x}} \, dx$

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

I think this was the best way to do it.
 

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