tentukanlah integral tentu berikut ini

Integral Tertentu Dgn substitusi

₀⁴∫ (2x +1) √(x + x²) dx =

u = x + x²
du = (2x +1) dx

u = x + x²
x= 4 --> u = 4 +16 = 20
x = 0 --> u = 0 + 0 = 0

₀⁴∫ (2x +1) √(x + x²) dx  = ₀²⁰∫ √ u d u = ₀²⁰∫ u^(`1/2) du

= [ 2/3   u^(3/2)]²⁰₀
= [2/3  u√u ]²⁰₀
= (2/3)  20√20
= (2/3) 20 (2√5)
= 80/3  √5