Soal UN MTK IPA 2016
Tentukan \( \displaystyle \int_{-1}^1 (2x^2-4x+3) \ dx \).
- \( \frac{22}{3} \)
- \( 6 \)
- \( \frac{16}{3} \)
- \( 4 \)
- \( \frac{4}{3} \)
Pembahasan:
Berdasarkan sifat integral tentu, diperoleh berikut:
\begin{aligned} \int_{-1}^1 (2x^2-4x+3) \ dx &= \left[ \frac{2}{3}x^3-\frac{4}{2}x^2+3x \right]_{-1}^1 \\[8pt] &= \left[ \frac{2}{3}x^3-2x^2+3x \right]_{-1}^1 \\[8pt] &= \left[ \frac{2}{3}(1)^3-2(1)^2+3(1) \right]-\left[ \frac{2}{3}(-1)^3-2(-1)^2+3(-1) \right] \\[8pt] &= \left[ \frac{2}{3}-2+3 \right]-\left[-\frac{2}{3}-2-3 \right] \\[8pt] &= \frac{2}{3}+1+\frac{2}{3}+5 \\[8pt] &= \frac{4}{3}+6 = \frac{4}{3}+\frac{18}{3} \\[8pt] &= \frac{22}{3} \end{aligned}
Jawaban A.