$\int\limits^2_0 {x*√(2x-x^2)} \, dx$

Ta tính

$I = \displaystyle \int_0^2 x\sqrt{2x-x^2}dx$

$= \displaystyle \int_0^2 (x-1) \sqrt{2x-x^2} dx + 2\displaystyle \int_0^2 \sqrt{2x-x^2}dx$

$= -\dfrac{1}{2} \displaystyle \int_0^2 \sqrt{2x-x^2} d(2x-x^2) + 2\displaystyle \int_0^2 \sqrt{2x-x^2}dx$

$= -\dfrac{1}{2} . \dfrac{2}{3} \sqrt{(2x-x^2)^3}\Bigg\vert_0^2 + 2\displaystyle \int_0^2 \sqrt{2x-x^2}dx$

$= 2\displaystyle \int_0^2 \sqrt{2x-x^2} dx$

Ta tính

$\displaystyle \int_0^2 \sqrt{2x-x^2} dx = \displaystyle \int_0^2 \sqrt{1 - (x-1)^2} dx$

Đặt $u = x-1$, suy ra cận chạy từ -1 đến 1

$= \displaystyle \int_{-1}^1 \sqrt{1 - u^2} du$

Đặt $u = \sin t$. Khi đó $du = \cos t dt$. Cận chạy từ $-\dfrac{\pi}{2}$ đến $\dfrac{\pi}{2}$. Tích phân trở thành

$=\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos t \sqrt{1 - \sin^2t} dt$

$= \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2t dt$

$= \dfrac{1}{2} \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2t)+1$

$= \dfrac{1}{2} \left( \sin(2t) + t \right) \Bigg\vert_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$= \dfrac{1}{2} .\pi$

Do đó

$I = 2. \dfrac{1}{2} \pi = \pi$.

Đáp án:

Giải thích các bước giải:

$\int _0^2\:x\cdot \sqrt{2x-x^2}dx$

$=\int _0^2x\sqrt{-\left(x-1\right)^2+1}dx$

Đặt `u=x-1`

$=\int _{-1}^1\left(u+1\right)\sqrt{-u^2+1}du$

$=\int _{-1}^1u\sqrt{-u^2+1}+\sqrt{-u^2+1}du$

$=\int _{-1}^1u\sqrt{-u^2+1}du+\int _{-1}^1\sqrt{-u^2+1}du$

$=\dfrac{\pi }{2}$