Integral (2x² + 4x – 3) batas atas 3 batas bawah 1 dx =

by

³₁

= [⅔(3³) + 2(3²) – 3(3)] – [⅔(1³) + 2(1²) – 3(1)]
= [18 + 18 – 9] – [⅔ + 2 – 3]
= 27 – ⅔ + 1
= 27⅓
  Pola Soal Limit Fungsi Aljabar Dan Pembahasannya