Soal SBMPTN 2018
Tentukan hasil dari \( \displaystyle \int_0^3 \frac{3x}{\sqrt{x+1}} \ dx \).
- 3
- 6
- 8
- 9
- 12
Pembahasan:
Kita bisa selesaikan soal ini menggunakan teknik substitusi. Misalkan \( u = x+1 \) sehingga diperoleh:
\begin{aligned} u = x+1 \Leftrightarrow x &= u-1 \\[8pt] \frac{dx}{du} &= 1 \\[8pt] dx &= du \end{aligned}
Selanjutnya, ganti batas dari integralnya, yakni:
\begin{aligned} x = 0 \Rightarrow u = x+1 = 0+1 = 1 \\[8pt] x = 3 \Rightarrow u = x+1 = 3+1 = 4 \end{aligned}
Substitusi hasil yang diperoleh di atas ke soal integral, diperoleh:
\begin{aligned} \int_0^3 \frac{3x}{\sqrt{x+1}} \ dx &= \int_1^4 \frac{3(u-1)}{\sqrt{u}} \ du = \int_1^4 \frac{3u-3}{\sqrt{u}} \ du \\[8pt] &= \int_1^4 \frac{3u}{\sqrt{u}} \ du - \int_1^4 \frac{3}{\sqrt{u}} \ du \\[8pt] &= \int_1^4 3u^{1/2} \ du - \int_1^4 3u^{-1/2} \ du \\[8pt] &= \left[ 3 \cdot \frac{2}{3}u^{3/2} \right]_1^4 - \left[ 3 \cdot 2u^{1/2} \right]_1^4 \\[8pt] &= \left[ 2u^{3/2} \right]_1^4 - \left[6u^{1/2} \right]_1^4 \\[8pt] &= (2(4)^{3/2}-2(1)^{3/2})-(6(4)^{1/2}-6(1)^{1/2}) \\[8pt] &= (16-2)-(12-6) \\[8pt] &= 8 \end{aligned}
Jawaban C.