Tentukan nilai integral berikut pake cara. Tolongg

semoga membantu yaa :)

Materi : Kalkulus
Bahasan : Integral Aljabar
Jawaban : 68

Langkah - langkah :

[tex] \int\limits^3_1 {(6x^2 + 4x)} \, dx \\ \\ = \frac{6x^{2 + 1}}{2 + 1} + \frac{4x^{1 + 1}}{1 + 1} \\ \\ = \frac{6x^3}{3} + \frac{4x^2}{2} \\ \\ = 2x^3 + 2x^2 \\ \\ Substitusi (3) \\ \\ = 2(3)^3 + 2(3)^2 \\ \\ = 54 + 18 = 72 \\ \\ Substitusi (1) \\ \\ = 2(1)^3 + 2(1)^2 \\ \\ = 2 + 2 = 4 \\ \\ Selisihkan \\ \\ = 72 - 4 \\ \\ = 68 [/tex]