Найти площадь фигуры, ограниченной линиями: y=arcsin2x, x=0, y= -π/2. Created by matwej2201. algebra-ru

2...

matwej2201 @matwej2201

July 2022 1 9 Report

Найти площадь фигуры, ограниченной линиями: y=arcsin2x, x=0, y= -π/2

Answers & Comments

NNNLLL54

Verified answer

Ответ:

$y=arcsin2x\ \ ,\ \ x=0\ \ ,\ \ y=-\dfrac{\pi}{2}\\\\-1\leq 2x\leq 1\ \ \to \ \ -\dfrac{1}{2}\leq x\leq \dfrac{1}{2}\ \ ,\ \ \ -\dfrac{\pi}{2}\leq arcsin2x\leq \dfrac{\pi }{2}\\\\\\S=\int\limits_0^{1/2}\Big(\dfrac{\pi}{2}-arcsin2x\Big)\, dx=\int \limits _0^{1/2}\dfrac{\pi}{2}\, dx-\int\limits_0^{1/2}\, arcsin2x\, dx=\\\\\\=\Big[u=arcsin2x\ ,\ du=\dfrac{2\, dx}{\sqrt{1-4x^2}}\ ,\ dv=dx\ ,\ v=x\ \Big]=$

$=\dfrac{\pi}{2}\cdot x\Big|^{1/2}_0-x\cdot arcsin2x\Big|^{1/2}_0+\int\limits_0^{1/2}\, \dfrac{2x\, dx}{\sqrt{1-4x^2}}=\Big[\ d(1-4x^2)=-8x\, dx\ \Big]=\\\\\\=\dfrac{\pi}{2}(0+\dfrac{1}{2})-\dfrac{1}{2}\cdot arcsin1-\dfrac{1}{4}\int\limits_0^{1/2}\, \dfrac{d(1-4x^2)}{\sqrt{1-4x^2}}dx=\\\\\\=\dfrac{\pi}{4}-\dfrac{\pi}{4}-\dfrac{1}{4}\cdot 2\sqrt{1-4x^2}\Big|^{1/2}_0=-\dfrac{1}{2}\cdot (0-\sqrt1)=\dfrac{1}{2}$

1 votes Thanks 0

Answers & Comments

Verified answer

pri szhiganii 22 g veshestva poluchili 44 g oksida ugleroda i 18 g vody otnosit

tex]...

sin168...

pryamaya u 4h11 parallelna kasatelnoj k grafiku funkcii u h 5h 6 najdit

tex]...

esse na temu moyo otnoshenie k prazdniku 4 noyabrya srochno pomogite pliz880d5cdf0320fe89f17f67e972a23430 32787

4 predstavit ugol dvumya sposobamic216a815b9ee9254a2f2c03e6cdfde4c 94036

4 predstavit ugol dvumya sposobami

4 obyasnite pozhalujsta9cc81df632e0c45a1c07091df25c34e2 5626

tex]...

рекомендуемые вопросы

Helpful Links

Helpful Social