Question

bantu jawab 3 soal ini segera ​

Answer 1

INTegral

i)

[tex]\sf \int {\dfrac{x^2}{\sqrt{x^3 + 1}}}dx[/tex]

[tex]\sf \int {x^2}(x^3 + 1)^{-\frac{1}{2}} dx[/tex]

u = x³ + 1
du =  3x² dx
x² dx  =  1/3 du

[tex]\sf \int {\dfrac{x^2}{\sqrt{x^3 + 1}}}dx= \sf \int {\frac{1}{3}}U^{-\frac{1}{2}}\ du[/tex]

[tex]\sf = \frac{1}{6}U^{\frac{1}{2}} + c = \frac{1}{6}\sqrt{(x^3 + 1)} +c[/tex]

2. ∫ sin³ 2θ dθ
= ∫ (sin 2θ)³ dθ
= ∫ (2 sin θ cos θ)³ dθ

= ∫ 8 sin³ θ cos³ θ dθ

= 8 ∫ sin² θ. sin θ. cos³  dθ

= 8 ∫ (1 - cos²θ) cos³θ. sin θ dθ
u = cos θ
du =  - sin θ dθ ⇒  sin θ dθ  = - du
= 8 ∫ (1 - cos²θ) cos³θ. sin θ dθ = 8 ∫ (1 -u²) u³  (- dθ)
= -8 ∫ (1 -u²) u³ dθ
= -8 ∫ (u³ - u⁵) dθ
= - 8 [ ¹/₄ u⁴ - ¹/₆ u⁶] + c
= - 2u⁴ + 4/3 u⁶ + c
= - 1/3 u⁴ (6- 4u²) + c
= - 1/3 cos⁴ θ  (6 - 4 cos²θ) + c

3.

[tex]\sf \int_{-1}^2|2x- 3| dx[/tex]
2x -3 = 0
x =  3/2

[tex]\sf = \int\limits^{\frac{3}{2}}_{-1} -(2 x+ 3) \, dx+ \int\limits^{2}_{\frac{3}{2}} +(2 x+ 3) \, dx[/tex]

[tex]\sf = \int\limits^{\frac{3}{2}}_{-1} -2 x- 3 \, dx+ \int\limits^{2}_{\frac{3}{2}} 2 x+ 3 \, dx[/tex]

[tex]\sf = [ -x^2 - 3x]^{\frac{3}{2}}_{-1} + [ x^2 + 3x]^{2}_{\frac{3}{2}}[/tex]

[tex]\sf = \frac{25}{4} + \frac{1}{4}= \frac{26}{4}[/tex]

[tex]\sf = \frac{13}{2}= 6, 5[/tex]