A: \(2\pi {a^{2n + 1}}\)
B: \(2\pi {a^{2n - 1}}\)
C: \(\pi {a^{2n + 1}}\)
D: \(\pi {a^{2n - 1}}\)
举一反三
- 计算\({\oint_L {({x^2} + {y^2})} ^n}ds\),其中\(L\)为圆周\(x = a\cos t\),\(y=asint\)\((0 \le t \le 2\pi )\)。 A: \(2\pi {a^{n + 1}}\) B: \(2\pi {a^{2n + 1}}\) C: \(\pi {a^{n + 1}}\) D: \(2\pi {a^{n + 1}}\)
- 计算曲线积分\({\oint_L {({x^2} + {y^2})} ^3}ds\),其中\(L\)为圆周\(x = a\cos t,y = a\sin t(0 \le t \le 2\pi )\)。 A: \(2\pi {a^7}\) B: \(2\pi {a^6}\) C: \(2\pi {a^5}\) D: \(2\pi {a^8}\)
- 函数\(f(x) = x^2,\; x \in [-\pi,\pi]\)的Fourier级数为 A: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) B: \(\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\) C: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]\) D: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)
- 曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$
- 中国大学MOOC: 下列程序的运行结果是( )。x=0:pi/100:2*pi;for n=1:2:10 plot(n*sin(x),n*cos(x)) hold onendaxis square
内容
- 0
旋轮线$x=a(t-\sin t),y=a(1-\cos t)$的一拱($0 \le t \le 2 \pi$)的绕$x$轴旋转得到的立体的体积为 A: $\pi a^3$ B: $\frac{32}{105} \pi a^3$ C: $\pi a^2$ D: $\frac{32}{105} \pi a^2$
- 1
\(已知二元函数f(x,y)=\sin{x^2y},则\frac{\partial f}{\partial x}(1,\pi)=(\,)\) A: \(\frac{\pi}{2}\) B: \(2\pi\) C: \(-2\pi\) D: \(-\frac{\pi}{2}\)
- 2
函数$f(x)=\arcsin(\sin x)$的傅里叶级数展开式为 A: $x$ B: $$\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^n\sin(2n+1)x}{(2n+1)^2}$$ C: $$\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n\sin(2n+1)x}{(2n+1)^2}$$ D: $$\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sin(2n+1)x}{(2n+1)^2}$$
- 3
计算\(\int_{\;L} {ydx + xdy} \),其中 \(L\)为圆周 \(x = R\cos t\), \(y = R\sin t\)上对应 \(t = 0\)到 \(t = {\pi \over 2}\)的一段弧。 A: -1 B: 1 C: 0 D: 2
- 4
\( \sin x \)的麦克劳林公式为( ). A: \( \sin x = x - { { {x^3}} \over {3!}} + { { {x^5}} \over {5!}} - \cdots + {( - 1)^n} { { {x^{2n + 1}}} \over {\left( {2n + 1} \right)!}} + o\left( { { x^{2n + 2}}} \right) \) B: \( \sin x = 1 - { { {x^2}} \over {2!}} + { { {x^4}} \over {4!}} - { { {x^6}} \over {6!}} + \cdots + {( - 1)^n} { { {x^{2n}}} \over {\left( {2n} \right)!}} + o\left( { { x^{2n + 1}}} \right) \) C: \( \sin x = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)