Nomor 9 dan 10 tolong bntuannya ya..trma kasih.... Question from @YSP

October 2018 1 29 Report

Nomor 9 dan 10 tolong bntuannya ya..trma kasih

Takamori37 Nomor 9.
$$\begin{align}g(x)&=\cos(f(x)) \\ g'(x)&=[\cos'(f(x))].[f'(x)] \\ g'(x)&=-\sin(f(x)).f'(x) \\ g'(0)&=-\sin(f(0)).f'(0) \\ &=-\sin(1)\times2 \\ &=-2.\sin(1)\ \ [A] \end{align}$

Nomor 10.
Andaikan:
x + 1 = u
Sehingga, dx = du
Menjadi:
$$\begin{align}\int \frac{x+1}{(x+2)^4}\, dx&=\int \frac{\boxed{u}\to U}{\boxed{(u+1)^4}\to dV}\, du \\ \text{Integrasikan }&\text{secara parsial.} \\ &=UV-\int VdU \\ &=-\frac{u}{3(u+1)^3}-\int \frac{-1}{3(u+1)^3}\, du \\ &=-\frac{u}{3(u+1)^3}-\frac{1}{6(u+1)^2}+C \\ &=\frac{-2u-(u+1)}{6(u+1)^3}+C \\ &=\frac{-(3u+1)}{6(u+1)^3}+C\end{align}$
Sehingga, integral berikut:
$$\begin{align}\int \frac{x+1}{(x+2)^4}\, dx&=\frac{-(3(x+1)+1)}{6((x+1)+1)^3}+C \\ &=\frac{-(3x+4)}{6(x+2)^3}+C\end{align}$
Selesaikan:
$$\begin{align}\int_{-1}^a \frac{x+1}{(x+2)^4}\, dx&=\left.\frac{-(3x+4)}{6(x+2)^3}\right]_{-1}^a \\ \frac{10}{81}&=\left[\frac{-(3a+4)}{6(a+2)^3}\right]-\left[\frac{-(3(-1)+4)}{6((-1)+2)^3}\right]\right] \\ \frac{10}{81}&=\frac{-(3a+4)}{6(a+2)^3}-\left[\frac{-1}{6}\right] \\ \frac{10}{81}-\frac16&=\frac{-(3a+4)}{6(a+2)^3} \\ -\frac7{162}&=\frac{-(3a+4)}{6(a+2)^3} \\ \frac7{27}&=\frac{3a+4}{(a+2)^3} \end{align}$
Jika diterka kembali, tentu a yang tepat adalah 1.
Sehingga, a = 1
- Cek kembali pilihannya -

1 votes Thanks 1

YSP Ketemu jawabannya D thanks banget, sangat membantu

Takamori37 Iya, sama-sama ~