Integral 2x + 2/√x² + 2x + 1 batas atas 4 batas bawah 0 dx

=

8
Pembahasan:
₀∫⁴      2x + 2      dx
      √x² + 2x + 1
= ₀∫⁴ (2x + 2)(x² + 2x + 1)⁻½ dx
Misal,
u = x² + 2x + 1
du = 2x + 2 dx, maka
₀∫⁴ (2x + 2)(x² + 2x + 1)⁻½ dx
= ₀∫⁴ u⁻½ du
= [2u½]⁴₀
= [2√x² + 2x + 1]⁴₀
= [2√4² + 2.4 + 1] – [2√0² + 2.0 + 1]
= 2√25 – 2√1
= 2.5 – 2.1
= 10 – 2
= 8
  Matriks x yang memenuhi x (-6 -5 5 4) = (2 1 4 3)