Jika \( \displaystyle \int_1^2 f(x) \ dx = \sqrt{2} \), maka nilai \( \displaystyle \int_1^4 \frac{1}{\sqrt{x}} f( \sqrt{x} ) \ dx \) adalah…
- \( \frac{\sqrt{2}}{4} \)
- \( \frac{\sqrt{2}}{2} \)
- \( \sqrt{2} \)
- \( 2\sqrt{2} \)
- \( 4\sqrt{2} \)
(SBMPTN 2018)
Pembahasan:
Untuk mengerjakan soal ini, kita perlu mengingat ini:
\begin{aligned} \int_a^b \ f(x) \ dx = \int_a^b f(u) \ du = \int_a^b f(t) \ dt = \cdots \end{aligned}
Sekarang, misalkan \(u = \sqrt{x} \) sehingga diperoleh:
\begin{aligned} u = \sqrt{x} \Leftrightarrow \frac{du}{dx} &= \frac{1}{2\sqrt{x}} \\[8pt] dx &= 2\sqrt{x} \ du \\[8pt] dx &= 2u \ du \end{aligned}
Selanjutnya, ganti batas dari integralnya, yakni:
\begin{aligned} x &= 1 \Rightarrow u = \sqrt{x} = \sqrt{1} = 1 \\[8pt] x &= 4 \Rightarrow u = \sqrt{x} = \sqrt{4} = 2 \end{aligned}
Berdasarkan hasil di atas, maka diperoleh berikut ini:
\begin{aligned} \int_1^4 \frac{1}{\sqrt{x}} \ f( \sqrt{x} ) \ dx &= \int_1^2 \frac{1}{u} f(u) \cdot 2u \ du \\[8pt] &= 2 \int_1^2 f(u) \ du \\[8pt] &= 2 \int_1^2 f(x) \ dx \\[8pt] &= 2\sqrt{2} \end{aligned}
Jawaban D.