Question

HARAP BANTUANNYA SOAL INTEGRAL

Answer 1

5)
₀∫² (3x +4)¹/² dx =
u = 3x + 4
du = 3 dx  --> dx = 1/3 du
.
batas u
x = 2 --> u = 10
x  = 0--> u = 4
= ₄∫¹⁰ ¹/₃ u ¹/² du 
= ¹/₃ .²/₃. u√u |¹⁰₂
= ²/₉ (10√10 - 2√2)

6)
(π..0) ∫( sin 2x . cos x) dx =
(π..0) ∫ 1/2 ( 2 sin 2x. cos x) dx
(π..0) ∫ 1/2 ( sin 3x + sin x)
= -1/6 cos 3x - 1/2 cos x | (π...0)
= -1/6 (cos 3π - cos 0) - 1/2 (cos π - cos 0 )
= -1/6 (-1 - 1) - 1/2 (-1  - 1)
= -1/6(-2)  - 1/2 (-2)
= 1/3 + 1
= 4/3

7)
₀∫¹ 3x √(3x² +1) dx
u = 3x² +1 
du = 6x dx --> 3xdx = 1/2 du
batas intgral
x= 1 --> u = 4
x = 0 --> u = 1

₁∫⁴   1/2 u¹/² du =
= [ 1/2 . 2/3 u³/² ]⁴₁
= 1/3 [ u√u]⁴₁
= 1/3 (4√4 - 1√1)
= 1/3 ( 8 - 1)
= 7/3