integralkanlah pake cara.. bantu dong.. pliss

Materi : Kalkulus
Bahasan : Integral Aljabar
Jawaban : 40

Langkah - langkah :
[tex] \int\limits^3_{-1} {(3x^2 + 4x - 1)} \, dx \\ \\ = \frac{3x^{2 + 1}}{2 + 1} + \frac{4x^{1 + 1}}{1 + 1} - x \\ \\ = x^3 + 2x^{2} - x \\ \\ Substitusi(3) \\ = (3)^3 + 2(3)^2 - 3 \\ \\ = 27 + 18 - 3 \\ \\ = 42 \\ \\ Substitusi (- 1) \\ = (-1)^3 + 2(- 1)^{2} - (- 1) \\ \\ = - 1 + 2 + 1 \\ \\ = 2 \\ \\ Selisihkan \\ = 42 - 2 \\ \\ = 40[/tex]