Question

Soal integral tentu subtitusi dan parsial mohon bantuannya ^^

Answer 1

1) ₋₁¹∫(x-1)(x² -2x+9)²
u = x² -2x + 9 --> u' = 2x -2 dx ---> (x-1) dx = 1/2 du
batas x = 1 --> u = 8
batas x = -1 --> u =12

₁₂⁸∫ 1/2 U² du = 1/6 U³ |⁸₁₂
1/6( 8³ -12³) = 1/6 ( - 1.216) = - 1216/6 = - 608/3

2)
(π/2 ...π/3)∫ x cos x dx =
[ x sin x + cos x |(π/2 ..π/3)
= (π/2 - π/3) (sin π/2 - sin π/3)  + cos π/2 - cos π/3
= π/6 ( 1- 1/2√3) + 0 - 1/2
= π/12 (2-√3) - 1/2

3)
π/2 ..π/3 [ -1/2 x cos 2x + 1/4 sin 2x]
= -1/2 (π/2 - π/3) { cos π - cos 2π/3) + 1/4 { sin π - sin 2π/3}
= - 1/12 π (-1 + - 1/2) + 1/2 (0 - 1/2√3)
= - 1/12 π (- 3/2) + 1/2(-1/2√3)
= 1/8 π + 1/4√3

Answer 2

1.

misal, x^2 -2x + 9 = u
2x - 2 dx = du
2(x - 1)dx = du
dx = du/(2(x - 1))

saya gak tulis batasnya, ya, tapi nanti jangan lupa tulis.

∫(x - 1)(x^2 - 2x + 9)^2 dx = ∫(x - 1)(u)^2 du/(2(x - 1))
= ∫u^2 du/2
= (1/2)(1/3)u^3
= (1/6)u^3

...................................1
= (1/6)(x^2 -2x + 9)^3 |
..................................-1

= (1/6)(1^2 - 2(1) + 9)^3 - (1/6)((-1)^2 - 2(-1) + 9)^3
= 512/6 - 1728/6
= -1216/6
= -608/3

==========

2.

misal u = x
du = dx
dv = cosx dx
v = sinx

uv - ∫vdu
= x.sinx - ∫sinx dx

........................π/2
= x.sinx + cosx |
........................π/3

= π/2.sin(π/2) + cos(π/2) - π/3.sin(π/3) - cos(π/3)
= π/2 - (√3/6)π - 1/2
= π/2 - 1/6 √3 π - 1/2

==========

3.

misal u = x
du = dx
dv = sin2x dx
v = -(1/2)cos(2x)

uv - ∫v du
= x.(-1/2)cos(2x) - ∫-(1/2)cos(2x) dx
= -(1/2)xcos(2x) + ∫(1/2)cos(2x) dx
= -(1/2)xcos(2x) + (1/4)sin(2x)

........................................π/2
= 1/2(-xcos(2x) + (1/2)sin(2x)) |
........................................π/3

= 1/2(-(π/2)cos(π) + (1/2)sin(π)) - 1/2(-(π/3)cos(2π/3) +(1/2) sin(2π/3))
= 1/2(π/2) - 1/2(π/6 + √3/4)
= π/4 - π/12 - √3/8
= 3π/12 - π/12 - √3/8
= 2π/12 - √3/8
= π/6 - (1/8) √3