[tex]_{0}\int {}^{8} ( \sqrt{144 - 2 {x}^{2} } -

Berikut ini adalah pertanyaan dari unknown pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

$_{0}\int {}^{8} ( \sqrt{144 - 2 {x}^{2} } - 12 + x)dx =$ ???

Help................

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Hasil dari $\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx$ adalah $\boldsymbol{4\sqrt{2}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ]-64}$ .

PEMBAHASAN

Integral merupakan operasi yang menjadi kebalikan dari operasi turunan/diferensial. Sehingga integral sering juga disebut sebagai antiturunan.

$f(x)=\int\limits {\left [ \frac{df(x)}{dx} \right ]} \, dx$

Substitusi trigonometri dapat digunakan untuk menyelesaikan integral dengan bentuk :

$\sqrt{a^2-x^2}~\to~substitusi~sin\theta=\frac{x}{a}$

$\sqrt{x^2-a^2}~\to~substitusi~sec\theta=\frac{x}{a}$

$\sqrt{x^2+a^2}~\to~substitusi~tan\theta=\frac{x}{a}$

DIKETAHUI

$\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx=$

DITANYA

Tentukan hasil integralnya.

PENYELESAIAN

$\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx$

$=\int\limits^8_0 {\sqrt{144-2x^2}} \, dx+\int\limits^8_0 {\left ( -12+x \right )} \, dx$

Misal

$A=\int\limits^8_0 {\sqrt{144-2x^2}} \, dx$

$A=\int\limits^8_0 {\sqrt{2(72-x^2)}} \, dx$

$A=\int\limits^8_0 {\sqrt{2}\sqrt{72-x^2}} \, dx$

Gunakan substitusi trigonometri.

$sin\theta=\frac{x}{6\sqrt{2}}$

$6\sqrt{2}sin\theta=x$

$6\sqrt{2}cos\theta d\theta=dx$

$A=\int\limits^8_0 {\sqrt{2}\sqrt{72-(6\sqrt{2}sin\theta)^2}} \, (6\sqrt{2}cos\theta d\theta)$

$A=12\int\limits^8_0 {cos\theta\sqrt{72(1-sin^2\theta)}} \, d\theta$

$A=12\int\limits^8_0 {cos\theta\sqrt{72cos^2\theta}} \, d\theta$

$A=12\sqrt{72}\int\limits^8_0 {cos^2\theta} \, d\theta$

$A=12\sqrt{72}\int\limits^8_0 {\left ( \frac{1}{2}+\frac{1}{2}cos2\theta \right )} \, d\theta$

$A=12\sqrt{72}\left ( \frac{1}{2}\theta+\frac{1}{4}sin2\theta \right )\Bigr|^8_0$

$A=3\sqrt{72}\left ( 2\theta+sin2\theta \right )\Bigr|^8_0$

$A=3\sqrt{72}\left ( 2arcsin\left ( \frac{x}{6\sqrt{2}} \right )+2sin\theta cos\theta \right )\Bigr|^8_0$

$A=3\sqrt{72}\left ( 2arcsin\left ( \frac{x}{6\sqrt{2}} \right )+2\left ( \frac{x}{6\sqrt{2}} \right )\left ( \frac{\sqrt{72-x^2}}{6\sqrt{2}} \right ) \right )\Bigr|^8_0$

$A=3\sqrt{72}\left [ 2arcsin\left ( \frac{8}{6\sqrt{2}} \right )+2\left ( \frac{8}{6\sqrt{2}} \right )\left ( \frac{\sqrt{72-8^2}}{6\sqrt{2}} \right )-\left ( 2arcsin\left ( \frac{0}{6\sqrt{2}} \right )+2\left ( \frac{0}{6\sqrt{2}} \right )\left ( \frac{\sqrt{72-0^2}}{6\sqrt{2}} \right ) \right ) \right ]$

$A=3\sqrt{72}\left [ 2arcsin\left ( \frac{4}{3\sqrt{2}} \right )+2\left ( \frac{4}{3\sqrt{2}} \right )\left ( \frac{2\sqrt{2}}{6\sqrt{2}} \right )-\left ( 0+(0)\left ( \frac{\sqrt{72}}{6\sqrt{2}} \right ) \right ) \right ]$

$A=18\sqrt{2}\left [ 2arcsin\left ( \frac{2\sqrt{2}}{3} \right )+\frac{4\sqrt{2}}{9} \right ]$

$A=18\sqrt{2}\times\frac{2}{9}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ]$

$A=4\sqrt{2}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ]$

Misal

$B=\int\limits^8_0 {(-12+x)} \, dx$

$B=-12x+\frac{1}{2}x^2\Bigr|^8_0$

$B=-12(8)+\frac{1}{2}(8)^2-[-12(0)+\frac{1}{2}(0)^2]$

$B=-64$

Maka :

$\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx$

$=\int\limits^8_0 {\sqrt{144-2x^2}} \, dx+\int\limits^8_0 {\left ( -12+x \right )} \, dx$

$=A+B$

$=4\sqrt{2}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ]-64$

KESIMPULAN

Hasil dari $\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx$ adalah $\boldsymbol{4\sqrt{2}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ]-64}$ .

PELAJARI LEBIH LANJUT

Integral substitusi trigonometri : yomemimo.com/tugas/40327197
Integral substitusi trigonometri : yomemimo.com/tugas/30251199
Integral substitusi trigonometri : yomemimo.com/tugas/30205263

DETAIL JAWABAN

Kelas : 11

Mapel: Matematika

Bab : Integral

Kode Kategorisasi: 11.2.10

Kata Kunci : integral, antiturunan, substitusi, trigonometri.

$Hasil dari [tex]\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx[/tex] adalah [tex]\boldsymbol{4\sqrt{2}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ]-64}[/tex].PEMBAHASANIntegral merupakan operasi yang menjadi kebalikan dari operasi turunan/diferensial. Sehingga integral sering juga disebut sebagai antiturunan. [tex]f(x)=\int\limits {\left [ \frac{df(x)}{dx} \right ]} \, dx[/tex]Substitusi trigonometri dapat digunakan untuk menyelesaikan integral dengan bentuk :[tex]\sqrt{a^2-x^2}~\to~substitusi~sin\theta=\frac{x}{a}[/tex] [tex]\sqrt{x^2-a^2}~\to~substitusi~sec\theta=\frac{x}{a}[/tex][tex]\sqrt{x^2+a^2}~\to~substitusi~tan\theta=\frac{x}{a}[/tex].DIKETAHUI[tex]\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx=[/tex].DITANYATentukan hasil integralnya..PENYELESAIAN[tex]\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx[/tex][tex]=\int\limits^8_0 {\sqrt{144-2x^2}} \, dx+\int\limits^8_0 {\left ( -12+x \right )} \, dx[/tex].Misal [tex]A=\int\limits^8_0 {\sqrt{144-2x^2}} \, dx[/tex] [tex]A=\int\limits^8_0 {\sqrt{144-2x^2}} \, dx[/tex][tex]A=\int\limits^8_0 {\sqrt{2(72-x^2)}} \, dx[/tex][tex]A=\int\limits^8_0 {\sqrt{2}\sqrt{72-x^2}} \, dx[/tex]Gunakan substitusi trigonometri.[tex]sin\theta=\frac{x}{6\sqrt{2}}[/tex][tex]6\sqrt{2}sin\theta=x[/tex][tex]6\sqrt{2}cos\theta d\theta=dx[/tex].[tex]A=\int\limits^8_0 {\sqrt{2}\sqrt{72-(6\sqrt{2}sin\theta)^2}} \, (6\sqrt{2}cos\theta d\theta)[/tex][tex]A=12\int\limits^8_0 {cos\theta\sqrt{72(1-sin^2\theta)}} \, d\theta[/tex][tex]A=12\int\limits^8_0 {cos\theta\sqrt{72cos^2\theta}} \, d\theta[/tex][tex]A=12\sqrt{72}\int\limits^8_0 {cos^2\theta} \, d\theta[/tex][tex]A=12\sqrt{72}\int\limits^8_0 {\left ( \frac{1}{2}+\frac{1}{2}cos2\theta \right )} \, d\theta[/tex][tex]A=12\sqrt{72}\left ( \frac{1}{2}\theta+\frac{1}{4}sin2\theta \right )\Bigr|^8_0[/tex][tex]A=3\sqrt{72}\left ( 2\theta+sin2\theta \right )\Bigr|^8_0[/tex][tex]A=3\sqrt{72}\left ( 2arcsin\left ( \frac{x}{6\sqrt{2}} \right )+2sin\theta cos\theta \right )\Bigr|^8_0[/tex][tex]A=3\sqrt{72}\left ( 2arcsin\left ( \frac{x}{6\sqrt{2}} \right )+2\left ( \frac{x}{6\sqrt{2}} \right )\left ( \frac{\sqrt{72-x^2}}{6\sqrt{2}} \right ) \right )\Bigr|^8_0[/tex][tex]A=3\sqrt{72}\left [ 2arcsin\left ( \frac{8}{6\sqrt{2}} \right )+2\left ( \frac{8}{6\sqrt{2}} \right )\left ( \frac{\sqrt{72-8^2}}{6\sqrt{2}} \right )-\left ( 2arcsin\left ( \frac{0}{6\sqrt{2}} \right )+2\left ( \frac{0}{6\sqrt{2}} \right )\left ( \frac{\sqrt{72-0^2}}{6\sqrt{2}} \right ) \right ) \right ][/tex][tex]A=3\sqrt{72}\left [ 2arcsin\left ( \frac{4}{3\sqrt{2}} \right )+2\left ( \frac{4}{3\sqrt{2}} \right )\left ( \frac{2\sqrt{2}}{6\sqrt{2}} \right )-\left ( 0+(0)\left ( \frac{\sqrt{72}}{6\sqrt{2}} \right ) \right ) \right ][/tex][tex]A=18\sqrt{2}\left [ 2arcsin\left ( \frac{2\sqrt{2}}{3} \right )+\frac{4\sqrt{2}}{9} \right ][/tex][tex]A=18\sqrt{2}\times\frac{2}{9}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ][/tex][tex]A=4\sqrt{2}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ][/tex]..Misal[tex]B=\int\limits^8_0 {(-12+x)} \, dx[/tex][tex]B=-12x+\frac{1}{2}x^2\Bigr|^8_0[/tex][tex]B=-12(8)+\frac{1}{2}(8)^2-[-12(0)+\frac{1}{2}(0)^2][/tex][tex]B=-64[/tex].Maka :[tex]\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx[/tex][tex]=\int\limits^8_0 {\sqrt{144-2x^2}} \, dx+\int\limits^8_0 {\left ( -12+x \right )} \, dx[/tex][tex]=A+B[/tex][tex]=4\sqrt{2}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ]-64[/tex].KESIMPULANHasil dari [tex]\int\limits^8_0 {\left ( \sqrt{144-2x^2}-12+x \right )} \, dx[/tex] adalah [tex]\boldsymbol{4\sqrt{2}\left [ 9arcsin\left ( \frac{2\sqrt{2}}{3} \right )+2\sqrt{2} \right ]-64}[/tex]..PELAJARI LEBIH LANJUTIntegral substitusi trigonometri : https://brainly.co.id/tugas/40327197Integral substitusi trigonometri : https://brainly.co.id/tugas/30251199Integral substitusi trigonometri : https://brainly.co.id/tugas/30205263.DETAIL JAWABANKelas : 11Mapel: MatematikaBab : IntegralKode Kategorisasi: 11.2.10Kata Kunci : integral, antiturunan, substitusi, trigonometri.$

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