Solve $ \int_{-1}^1\frac{\sqrt{1-x^2}}{\pi}dx=$ :
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举一反三
- 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: $\frac{\pi}{4}$.
- Solve $ \frac{1}{\pi}\int_0^{\frac{\pi}{2}}\sin^4{x}dx=$ :<br/>______
- 4. $\int_{0}^{1}{\frac{{{e}^{x}}}{\sqrt{1-x}}dx}$的敛散性为( )。 (填“收敛”或“发散”)<br/>______
- \( \int_0^1 {dx} \int_ { { x^2}}^x { { {\left( { { x^2} + {y^2}} \right)}^{ - {1 \over 2}}}dy} \) =（ ） A: \( \sqrt 2 + 1 \) B: \( \sqrt 2 - 1 \) C: \( \sqrt 2 \) D: \( \pi \)
- \(已知曲线弧L:y=\sqrt{1-x^2}(0\le x\le 1).则\int_{L}xyds=(\,)\) A: \[1\] B: \[\frac{1}{2}\] C: \[\frac{1}{3}\] D: \[\frac{1}{4}\]