soal integral tentu mohon bantu, makasih

0∫6 (f(x)) dx
= 0∫3 (f(x)) dx + 3∫6 (f(x)) dx
= 0∫3 (x + 4) dx + 3∫6 (2 - 4x) dx
= [1/2 x^2 + 4x] (0|3) + [2x - 2x^2] (3|6)
= [(1/2 (3)^2 + 4(3)) - (1/2 (0)^2 + 4(0))] + [(2(6) - 2(6)^2) - (2(3) - 2(3)^2)]
= [(9/2 + 12) - (0 + 0)] + [(12 - 72) - (6 - 18)]
= 4,5 + 12 + (-60) - (-12)
= -31,5

= -31 1/2

= -63/2

integral tertentu
₀⁶ ∫ f (x) dx = ₀³ ∫ f(x) dx + ₃⁶∫ f(x) dx

=₀³ ∫ (x+4) dx + ₃⁶∫ 2 - 4x dx
= [¹/₂ x² + 4x]³₀  + [2x - 2x²]⁶₃
= 1/2(9-0) + 4(3-0) + 2(6-3) -  2(36-9)
= 9/2 + 12 + 6 - 27/2
= 18 - 18/2
= 9