Question

Quizz


Nilai dari
[tex] \displaystyle \int_{0}^{1} x \sqrt{x \sqrt[3]{x \sqrt[4]{x \sqrt[5]{...} } } } \: dx[/tex]
adalah
[tex]a. \: 2e[/tex]
[tex]b. \: e[/tex]
[tex]c. \: \frac{1}{e} [/tex]
[tex]d. \: \frac{1}{2e} [/tex]
[tex]e. \: \sqrt{e} [/tex]
[no ngasal, pakai cara]
[jawaban ada di opsi]


​

Answer 1

Pertama, kita manipulasi bentuk integral tak wajarnya

[tex] \begin{align}& \int_0^1 x \sqrt{x \sqrt[3]{x \sqrt[4]{x \sqrt[5]{...} } } } \: \text{ d}x \\ &= \int_0^1 x \cdot\sqrt{x} \cdot \sqrt{\sqrt[3]{x}} \cdot \sqrt{\sqrt[3]{\sqrt[4]{x}}} \cdots \: \text{ d}x \\ &= \int_0^1 x \cdot x^{\frac12}\cdot x^{\frac{1}{3\cdot2}}\cdot x^{\frac{1}{4\cdot3\cdot2}}\cdots \text{ d}x \\ &= \int_0^1 x^{\frac{1}{1!}}\cdot x^{\frac{1}{2!}}\cdot x^{\frac{1}{3!}} \cdot x^{\frac{1}{4!}}\cdots \text{ d}x \\ &=\int_0^1 x^{\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...} \text{ d}x \end{align} [/tex]

Bentuk [tex] \displaystyle \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+... = \sum_{n=1}^{\infty} \frac{1}{n!} [/tex].

Deret tersebut mirip dengan Deret Faktorial Terbalik:

[tex]\displaystyle \boxed{ \sum_{n=0}^{\infty}\frac{1}{n!} = e} [/tex]

Manipulasi sedikit deretnya menjadi

[tex] \begin{align} \sum_{n=0}^{\infty}\frac{1}{n!} &= e \\ \dfrac{1}{0!} + \sum_{n=1}^{\infty}\frac{1}{n!} &= e \\ 1+ \sum_{n=1}^{\infty}\frac{1}{n!} &= e \\ \sum_{n=1}^{\infty}\frac{1}{n!} &= e -1 \end{align} [/tex]

Maka, bentuk integralnya menjadi

[tex] \begin{align} \int_0^1 x^{e-1} \text{ d}x &= \left[ \dfrac{x^e}{e}\right]_{\blue0}^{\blue1} \\ &= \left[\dfrac{\blue1^e}{e}\right]-\left[\dfrac{\blue0^e}{e}\right] \\ &= \dfrac{1}{e}\end{align} [/tex]

JAWABAN YANG TEPAT ADALAH C.