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6 tháng 9 2021 lúc 22:51

\(I=\int\limits^3_0\sqrt{4-\left(x-2\right)^2}dx-\int\limits^3_0\sqrt{x}dx=\int\limits^3_0\sqrt{4-\left(x-2\right)^2}dx-2\sqrt{3}\)

Xét \(I_1=\int\limits^3_0\sqrt{4-\left(x-2\right)^2}dx\)

Đặt \(x-2=2sint\Rightarrow dx=2cost.dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=-\dfrac{\pi}{2}\\x=3\Rightarrow t=\dfrac{\pi}{6}\end{matrix}\right.\)

\(I_1=\int\limits^{\dfrac{\pi}{6}}_{-\dfrac{\pi}{2}}\sqrt{4-4sin^2t}.2cost.dt=4\int\limits^{\dfrac{\pi}{6}}_{-\dfrac{\pi}{2}}cos^2tdt=\int\limits^{\dfrac{\pi}{6}}_{-\dfrac{\pi}{2}}\left(2+2cos2t\right)dt\)

\(=\left(2t+sin2t\right)|^{\dfrac{\pi}{6}}_{-\dfrac{\pi}{2}}=\dfrac{4\pi}{3}+\dfrac{\sqrt{3}}{2}\)

\(\Rightarrow I=\dfrac{4\pi}{3}+\dfrac{\sqrt{3}}{2}-2\sqrt{3}=\dfrac{4\pi}{3}-\dfrac{3\sqrt{3}}{2}\)

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