Đáp án:
$\dfrac{2\pi}{3} +\dfrac{\sqrt3}{4}$
Giải thích các bước giải:
$\quad I =\displaystyle\int\limits_1^2x^2\sqrt{4 - x^2}dx$
Đặt $x = 2\sin u$
$\to dx = 2\cos udu$
Đổi cận:
$x\quad \Big|\quad 1\qquad 2$
$\overline{u\quad\Big|\quad \dfrac{\pi}{6}\quad \dfrac{\pi}{2}}$
Ta được:
$\quad I = \displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{2}}4\sin^2u\sqrt{4-4\sin^2u}.\cos udu$
$\to I = \displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{2}}16\sin^2u\cos^2udu$
$\to I = \displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{2}}4\sin^22udu$
$\to I = 4\displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{2}}\dfrac{1 - \cos4u}{2}du$
$\to I = 2\displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{2}}du - 2\displaystyle\int\limits_{\tfrac{\pi}{6}}^{\tfrac{\pi}{2}}cos4udu$
$\to I = 2u\Bigg|_{\tfrac{\pi}{6}}^{\tfrac{\pi}{2}} - \dfrac12\sin4u\Bigg|_{\tfrac{\pi}{6}}^{\tfrac{\pi}{2}}$
$\to I = \dfrac{2\pi}{3} +\dfrac{\sqrt3}{4}$
