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: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)
举一反三
- 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: $$\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sin(2n+1)x}{(2n+1)^2}$$
- 函数$f(x)=\sin x + \cos x,x \in [0,2 \pi]$的上凸区间为 A: $[0,\frac{\pi}{4}] \cup [\frac{5}{4} \pi,2 \pi] $ B: $[\frac{\pi}{4},\frac{5}{4} \pi]$ C: $[0,\frac{3}{4}\pi] \cup [\frac{7}{4} \pi,2 \pi] $ D: $[\frac{3}{4} \pi,\frac{7}{4} \pi] $
- 函数$f(x) =sin^3 x, x \in [0,2 \pi]$的单调递减区间为 A: $[\frac{\pi}{2},\frac{3}{2} \pi]$ B: $[\frac{3}{2} \pi,2 \pi]$ C: $[0,\frac{\pi}{2}]$ D: $[0,2 \pi]$
- 积分$\int_0^1 x \arctan xdx=$（）。 A: $\frac{\pi}{4}+\frac{1}{2}$ B: $\frac{\pi}{4}$ C: $\frac{\pi}{4}-\frac{1}{2}$ D: $\frac{1}{2}$
- \(已知二元函数f(x,y)=\sin{x^2y},则\frac{\partial f}{\partial x}(1,\pi)=(\,)\) A: \(\frac{\pi}{2}\) B: \(2\pi\) C: \(-2\pi\) D: \(-\frac{\pi}{2}\)
内容
- 0
Solve $ \frac{1}{\pi}\int_0^{\frac{\pi}{2}}\sin^4{x}dx=$ :<br/>______
- 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: $\frac{\pi}{4}$.
- 2
Solve $n \in \mathbb{N}, \int_0^{\frac{\pi}{2}}(\sin^n{x}-\cos^n{x})dx=$ :<br/>______
- 3
\(已知曲面\Sigma:x^2+y^2+z^2=a^2被平面z=h(0 A: \[2\pi a \ln\frac{a}{h}\] B: \[3\pi a \ln\frac{a}{h}\] C: \[4\pi a \ln\frac{a}{h}\] D: \[\pi a \ln\frac{a}{h}\]
- 4
8. 下列不等式正确的是 A: $0\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}\lt \int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}$ B: $0\lt \int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}$ C: $\int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}\lt 0\lt \int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}$ D: $\int_{0}^{\frac{\pi }{2}}{\cos (\sin x)dx}\lt 0\lt \int_{0}^{\frac{\pi }{2}}{\sin (\sin x)dx}$