∫²dan batas bawah 1 (Idk bagimana menulisnya) t(t-2) (t-1) dt Jawaban integral jalan dan penjelasan

Materi : Kalkulus
Bahasan : Integral Aljabar
Jawaban : [tex]- \frac{1}{4}[/tex]

Langkah - langkah :
t(t - 2) (t - 1) = t(t² - 3t + 2)
t(t - 2) (t - 1) = t³ - 3t² + 2t

[tex] \int\limits^2_1 {(t^{3} - 3t^{2} + 2t)} \, dt \\ \\ = \frac{t^{3 + 1}}{3 + 1} - \frac{3t^{2 + 1}}{2 + 1} + \frac{2t^{1 + 1}}{1 + 1} \\ \\ = \frac{1}{4}t^{4} - t^{3} + t^{2} \\ \\ Substitusi(2) \\ \\ = \frac{1}{4}(2^{4} ) - (2)^3 + (2)^2 \\ \\ = 4 - 8 + 4 = 0 \\ \\ Substitusi(1) \\ \\ = \frac{1}{4}(1^{4} ) - (1)^3 + (1)^2 \\ \\ = \frac{1}{4} - 1 + 1 = \frac{1}{4} \\ \\ Selisih \\ \\ 0 - \frac{1}{4} = - \frac{1}{4} [/tex]