\(已知曲线弧L:y=\sqrt{1-x^2}(0\le x\le 1).则\int_{L}xyds=(\,)\)
A: \[1\]
B: \[\frac{1}{2}\]
C: \[\frac{1}{3}\]
D: \[\frac{1}{4}\]
A: \[1\]
B: \[\frac{1}{2}\]
C: \[\frac{1}{3}\]
D: \[\frac{1}{4}\]
举一反三
- \(已知L为抛物线y^2=x上从点A(1,-1)到点B(1,1)的一段弧,则\int_{L}xyds=(\,)\) A: \[\frac{4}{5}\] B: \[\frac{3}{5}\] C: \[\frac{2}{5}\] D: \[\frac{1}{5}\]
- \(已知L是抛物线y=x^2上点O(0,0)与点A(1,1)之间的一段弧,则\int_{L}\sqrt{y}ds=(\,)\) A: \[\frac{1}{12}(5\sqrt{5}-1)\] B: \[\frac{1}{12}(3\sqrt{3}-1)\] C: \[\frac{1}{13}(5\sqrt{5}-1)\] D: \[\frac{1}{13}(3\sqrt{3}-1)\]
- Solve $\int_{-\frac{1}{2}}^1{1-x^2}dx=$? A: $\frac{\pi}{3}+\frac{\sqrt{3}}{8}$. B: $\frac{\pi}{2}$. C: $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$. D: $\frac{\pi}{4}$.
- 微分方程$y' = \sqrt{x},y(1)=0$的解为 A: $ \frac{2}{3} x^{\frac{3}{2}} + C $ B: $ \frac{2}{3} x^{\frac{3}{2}} -\frac{2}{3} $ C: $ x^{\frac{3}{2}}-1 $ D: $ x^{\frac{3}{2}}+C $
- 已知函数由下列方程确定$x^2 - y^2=1 $，则$\frac{d^2 y}{d^2 x} =$( )。 A: $\frac{1}{y^2}$ B: $-\frac{1}{y^2}$ C: $-\frac{1}{y^3}$ D: $\frac{1}{y^3}$