Hasil dari [tex]\lim_{x \to 0} \frac{\int\limits^x_0 {\sqrt{1+\text{sin} t} } \, dt }{x}[/tex] adala.... Question from @anginanginkel

$\displaystyle\lim_{x\to0}\dfrac{\int\limits_{0}^{x}{\sqrt{1+\sin t}}\,dt}{x}\ =\ \bf 1$

(opsi D)

Pembahasan

$\begin{aligned}&\lim_{x\to0}\frac{\displaystyle\int_{0}^{x}{\sqrt{1+\sin t}}\,dt}{x}\\\\&{\quad}\left[\ \begin{aligned}&{(\sin\alpha+\cos\alpha)}^2=1+\sin2\alpha\\&\therefore\ 1+\sin t=\left(\sin\left(\tfrac{t}{2}\right)+\cos\left(\tfrac{t}{2}\right)\right)^2\end{aligned}\right.\end{aligned}$

 
$\begin{aligned}&{=\ }\lim_{x\to0}\frac{\displaystyle\int_{0}^{x}{\sqrt{\left(\sin\left(\tfrac{t}{2}\right)+\cos\left(\tfrac{t}{2}\right)\right)^2}}\,dt}{x}\\&{=\ }\lim_{x\to0}\frac{\displaystyle\int_{0}^{x}{\left(\sin\left(\tfrac{t}{2}\right)+\cos\left(\tfrac{t}{2}\right)\right)}dt}{x}\\\\&{\quad}\left[\ \begin{aligned}&u=\frac{t}{2}\ \Rightarrow\ t=2u\\&\rightsquigarrow dt=2du\\&\rightsquigarrow\textsf{Batas atas }\Rightarrow\ \frac{x}{2} \end{aligned}\right.\end{aligned}$

 
$\begin{aligned}&{=\ }\lim_{x\to0}\frac{\displaystyle\int_{0}^{x/2}{(\sin u+\cos u)}\,2du}{x}\\&{=\ }\lim_{x\to0}\frac{\displaystyle2\int_{0}^{x/2}{(\sin u+\cos u)}\,du}{x}\\&{=\ }2\cdot\lim_{x\to0}\frac{\displaystyle\int_{0}^{x/2}{(\sin u+\cos u)}\,du}{x}\\&{=\ }2\cdot\lim_{x\to0}\frac{\displaystyle\int_{0}^{x/2}\sin u\,du+\int_{0}^{x/2}\cos u\,du}{x}\\&{=\ }2\cdot\lim_{x\to0}\frac{\bigl[-\cos u\bigr]_{0}^{x/2}+\bigl[\sin u\bigr]_{0}^{x/2}}{x}\end{aligned}$
$\begin{aligned}&{=\ }2\cdot\lim_{x\to0}\frac{\left[-\cos\left(\frac{x}{2}\right)-(-\cos0)\right]+\left[\sin\left(\frac{x}{2}\right)-\sin0\right]}{x}\\&{=\ }2\cdot\lim_{x\to0}\frac{-\cos\left(\frac{x}{2}\right)+1+\sin\left(\frac{x}{2}\right)}{x}\\&{=\ }2\cdot\lim_{x\to0}\frac{1+\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)}{x}\\\\&{\quad}\rightsquigarrow\textsf{bentuk tak tentu, karena 0/0}\\&{\quad}\rightsquigarrow\textsf{gunakan aturan L'H\^opital}\end{aligned}$

 
$\begin{aligned}&{=\ }2\cdot\lim_{x\to0}\frac{\frac{d}{dx}\left(1+\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)}{\frac{d}{dx}x}\\&{=\ }2\cdot\lim_{x\to0}\frac{\left(\cos\left(\frac{x}{2}\right)\right)\cdot\frac{d}{dx}\left(\frac{x}{2}\right)-\left(-\sin\left(\frac{x}{2}\right)\right)\cdot\frac{d}{dx}\left(\frac{x}{2}\right)}{1}\\&{=\ }2\cdot\lim_{x\to0}\left(\frac{1}{2}\cos\left(\frac{x}{2}\right)+\frac{1}{2}\sin\left(\frac{x}{2}\right)\right)\end{aligned}$
$\begin{aligned}&{=\ }\cancel{2}\cdot\frac{1}{\cancel{2}}\cdot\lim_{x\to0}\left(\cos\left(\frac{x}{2}\right)+\sin\left(\frac{x}{2}\right)\right)\\&{=\ }\lim_{x\to0}\left(\cos\left(\frac{x}{2}\right)+\sin\left(\frac{x}{2}\right)\right)\\&{=\ }\cos\left(\frac{0}{2}\right)+\sin\left(\frac{0}{2}\right)\\&{=\ }1+0\\&{=\ }\boxed{\ \bf1\ }\\\\&\blacksquare\end{aligned}$

5 votes Thanks 3

anginanginkel wah thank you banget kakk
kirain bakalan pake teorema gitu kak, tapi cara yg ini bisa dipahami bgt kak

henriyulianto saya nggak hafal teorema2. harus cari2 dulu di buku atau gugling. maklum kak, saya bukan ahlinya :D

anginanginkel aman kak, malah kalau pake teorema, nanti malah muncul yg aneh" teoremanya, harus dipelajarin dulu, ribet :(

diradiradira

Hasil dari $\displaystyle{ \lim_{x \to 0} \frac{\int\limits^x_0 {\sqrt{1+sint}} \, dt }{x} }$ adalah D. 1.

PEMBAHASAN

Integral merupakan operasi yang menjadi kebalikan dari operasi turunan/diferensial. Sehingga integral sering juga disebut sebagai antiturunan. Sifat - sifat operasi pada integral adalah sebagai berikut :

$(i)~\displaystyle{\int\limits {ax^n} \, dx=\frac{a}{n+1}x^{n+1}+C},~~~dengan~C=konstanta$

$(ii)~\displaystyle{\int\limits {kf(x)} \, dx=k\int\limits {f(x)} \, dx}$

$(iii)~\displaystyle{\int\limits {\left [ f(x)\pm g(x) \right ]} \, dx=\int\limits {f(x)} \, dx\pm\int\limits {g(x)} \, dx}$

$(iv)~\displaystyle{\int\limits^b_a {f(x)} \, dx=F(b)-F(a)}$

Sedangkan untuk mencari limit fungsi trigonometri dapat menggunakan rumus berikut :

$\displaystyle{(i)~\lim\limits_{x \to 0} \frac{sinax}{bx}=\lim\limits_{x \to 0} \frac{tanax}{bx}=\frac{a}{b} }$

$\displaystyle{(ii)~\lim\limits_{x \to 0} \frac{ax}{sinbx}=\lim\limits_{x \to 0} \frac{ax}{tanbx}=\frac{a}{b} }$

$\displaystyle{(iii)~\lim\limits_{x \to 0} \frac{sinax}{sinbx}=\lim\limits_{x \to 0} \frac{tanax}{tanbx}=\frac{a}{b} }$

$\displaystyle{(iv)~\lim\limits_{x \to a} \frac{sin(x-a)}{(x-a)}=\lim\limits_{x \to a} \frac{tan(x-a)}{(x-a)}=1 }$

DIKETAHUI

$\displaystyle{ \lim_{x \to 0} \frac{\int\limits^x_0 {\sqrt{1+sint}} \, dt }{x}= }$

DITANYA

Tentukan hasilnya.

PENYELESAIAN

Kita cari dahulu nilai dari $\displaystyle{\int\limits^x_0 {\sqrt{1+sint}} \, dt }$ .

$\displaystyle{\int\limits^x_0 {\sqrt{1+sint}} \, dt }$

$\displaystyle{=\int\limits^x_0 {\sqrt{1+sint}\times\frac{\sqrt{1-sint}}{\sqrt{1-sint}}} \, dt }$

$\displaystyle{=\int\limits^x_0 {\frac{\sqrt{1-sin^2t}}{\sqrt{1-sint}}} \, dt }$

$\displaystyle{=\int\limits^x_0 {\frac{\sqrt{cos^2t}}{\sqrt{1-sint}}} \, dt }$

$\displaystyle{=\int\limits^x_0 {\frac{|cost|}{\sqrt{1-sint}}} \, dt,~asumsi~x > 0 }$

$\displaystyle{=\int\limits^x_0 {\frac{cost}{\sqrt{1-sint}}} \, dt}$

$----------------$

Misal :

$u=1-sint~\to~du=-costdt$

$----------------$

$\displaystyle{=\int\limits^x_0 {\frac{cost}{\sqrt{u}}} \, \left ( \frac{du}{-cost} \right )}$

$\displaystyle{=-\int\limits^x_0 {u^{-\frac{1}{2}}} \, du}$

$\displaystyle{=-\left ( 2u^{\frac{1}{2}} \right )\Bigr|^x_0}$

$\displaystyle{=-2\sqrt{1-sint}\Bigr|^x_0}$

$\displaystyle{=-2\left ( \sqrt{1-sinx}-\sqrt{1-sin0} \right )}$

$\displaystyle{=-2\left ( \sqrt{1-sinx}-1 \right )}$

Sehingga :

$\displaystyle{ \lim_{x \to 0} \frac{\int\limits^x_0 {\sqrt{1+sint}} \, dt }{x}}$

$\displaystyle{=\lim_{x \to 0} \frac{-2\left ( \sqrt{1-sinx}-1 \right )}{x}}$

$\displaystyle{=-2\lim_{x \to 0} \frac{\sqrt{1-sinx}-1}{x}\times\frac{\sqrt{1-sinx}+1}{\sqrt{1-sinx}+1}}$

$\displaystyle{=-2\lim_{x \to 0} \frac{1-sinx-1}{x(\sqrt{1-sinx}+1)}}$

$\displaystyle{=-2\lim_{x \to 0} \frac{-sinx}{x(\sqrt{1-sinx}+1)}}$

$\displaystyle{=2\lim_{x \to 0} \frac{sinx}{x}\times\lim_{x \to 0} \frac{1}{\sqrt{1-sinx}+1}}$

$\displaystyle{=2\times1\times\frac{1}{\sqrt{1-sin0}+1}}$

$\displaystyle{=2\times\frac{1}{2}}$

$=1$

KESIMPULAN

Hasil dari $\displaystyle{ \lim_{x \to 0} \frac{\int\limits^x_0 {\sqrt{1+sint}} \, dt }{x} }$ adalah D. 1.

PELAJARI LEBIH LANJUT

Integral substitusi : brainly.co.id/tugas/47510453
Integral substitusi trigonometri : brainly.co.id/tugas/40327197
Integral substitusi trigonometri : brainly.co.id/tugas/30251199

DETAIL JAWABAN

Kelas : 11

Mapel: Matematika

Bab : Integral

Kode Kategorisasi: 11.2.10

3 votes Thanks 2

anginanginkel wahh thank you banget kakk
kak, mau nanya, kenapa pas asumsi x>0, |cos t| jadi cos t ya?
sama yg bagian pemisalan integral substitusi, nilai batasnya ngga berubah ka?

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PEMBAHASAN

DIKETAHUI

DITANYA

PENYELESAIAN

KESIMPULAN

PELAJARI LEBIH LANJUT

DETAIL JAWABAN

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