Question

Tolong bgt di bantu pr saya yaaaa please

Answer 1

Semoga membantuuuuuu.......

Answer 2

Integral (3x^2 - 3x + 4) (dx) dari 1 sampai 3
= x^3 - (3/2)(x^2) + 4x ] dari 1 smp 3
= [(3^3) - (3/2)(3^2) + 4(3)] - [(1^3) - (3/2)(1^2) + 4(1) ]
= (27 - (27/2) + 12) - (1 - (3/2) + 4)
= 39 - 27/2 - 5 + (3/2)
= 34 - (24/2)
= 34 - 12
= 22



integral (x-1)(3x+5)(dx) dari (-1) sampai 3
= integral (4x^2 - 3x + 5x -5) dari (-1) sampai 3
= integral (4x^2 + 2x - 5) dari (-1) sampai 3
= (4/3)(x^3) + (x^2) - 5x ] dari (-1) sampai 3
= [(4/3)(3^3) + (3^2) - 5(3)] - [(4/3)(-1)^3 + (-1)^2 - 5(-1)]
= (36 + 9 -15) - ((-4/3) + 1 + 5)
= 30 + (4/3) - 6
= 24 + (4/3)
= (72/3) + (4/3)
= 76/3