Let f, be a continuous function in [0, 1], then $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 0}^n {{1 \over n}} f\left( {{j \over n}} \right)$$ is

Let f be a differentiable function with $$\mathop {\lim }\limits_{x \to \infty } f(x) = 0.$$ If $$y' + yf'(x) - f(x)f'(x) = 0$$, $$\mathop {\lim }\limits_{x \to \infty } y(x) = 0$$, then (where $$y \equiv {{dy} \over {dx}})$$

If $$x\sin \left( {{y \over x}} \right)dy = \left[ {y\sin \left( {{y \over x}} \right) - x} \right]dx,\,x > 0$$ and $$y(1) = {\pi \over 2}$$, then the value of $$\cos \left( {{y \over x}} \right)$$ is

Let $$f(x) = 1 - \sqrt {({x^2})} $$, where the square root is to be taken positive, then