Toán $\int\limits^2_0 {x*√(2x-x^2)} \, dx$ 12/07/2021 By Josephine $\int\limits^2_0 {x*√(2x-x^2)} \, dx$
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` $=\int _{-1}^1\left(u+1\right)\sqrt{-u^2+1}du$ $=\int _{-1}^1u\sqrt{-u^2+1}+\sqrt{-u^2+1}du$ $=\int _{-1}^1u\sqrt{-u^2+1}du+\int _{-1}^1\sqrt{-u^2+1}du$ $=\dfrac{\pi }{2}$ Trả lời
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 $I = 2. \dfrac{1}{2} \pi = \pi$. Trả lời
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`
$=\int _{-1}^1\left(u+1\right)\sqrt{-u^2+1}du$
$=\int _{-1}^1u\sqrt{-u^2+1}+\sqrt{-u^2+1}du$
$=\int _{-1}^1u\sqrt{-u^2+1}du+\int _{-1}^1\sqrt{-u^2+1}du$
$=\dfrac{\pi }{2}$
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
$I = 2. \dfrac{1}{2} \pi = \pi$.