Question

[Kiến Văn]

Giúp mình với mn

Answer 1

[Nguyễn Việt Lâm]

1.

\(\sqrt{x}=0\Rightarrow x=0\)

\(\Rightarrow V=\pi\int\limits^1_0\left(\sqrt{x}\right)^2dx=\pi\)

2.

Phương trình hoành độ các giao điểm:

\(x^3=0\Rightarrow x=0\)

\(2-x=0\Rightarrow x=2\)

\(x^3=2-x\Rightarrow x=1\)

\(\Rightarrow S=\int\limits^1_0\left|x^3\right|dx+\int\limits^2_1\left|2-x\right|dx=\int\limits^1_0x^3dx+\int\limits^2_1\left(2-x\right)dx\)

\(=\dfrac{1}{2}+\int\limits^1_0x^3dx\)

Answer 2

[Nguyễn Việt Lâm]

3.

\(\dfrac{x^2}{4}+\dfrac{y^2}{1}=1\Rightarrow y^2=1-\dfrac{x^2}{4}\)

\(\Rightarrow y=\pm\sqrt{1-\dfrac{x^2}{4}}\)

Hoành độ giao điểm: \(\sqrt{1-\dfrac{x^2}{4}}=0\Rightarrow x=\pm2\)

\(\Rightarrow S=2\int\limits^2_{-2}\sqrt{1-\dfrac{x^2}{4}}dx\)

Đặt \(x=2sint\Rightarrow dx=2cost.dt\)

\(\left\{{}\begin{matrix}x=-2\Rightarrow t=-\dfrac{\pi}{2}\\x=2\Rightarrow t=\dfrac{\pi}{2}\end{matrix}\right.\)

\(\Rightarrow S=2\int\limits^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}\sqrt{1-cos^2t}.2cost.dt=2\int\limits^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}2cos^2tdt\)

\(=2\int\limits^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}\left(cos2t+1\right)dt=\left(sin2t+2t\right)|^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}}=2\pi\)