Integral batas bawah -2 batas atas 5 2x (3x+1) dx

By Admin ReiPosted on December 6, 2022

Integral batas bawah -2 batas atas 5 2x (3x+1) dx

Jawab:

$sf int_{-2}^5 ~ 2x(3x+ 1) dx$

$sf = int_{-2}^5 ~ (6x^2 + 2x ) dx$

$sf = 2x^3 + x^2 ]_{-2}^{5}$

$sf = 2(5^3 -(-2)^3) + (5^2-(-2)^2)$

$sf = 2(125+8) + (25-4)$

$sf = 266 - 21 = 245$