The number of continuous functions $f:[0,1] \rightarrow(-\infty, \infty)$ satisfying the condition $\int \limits_0^1(f(x))^2 dx =2 \int_0^1 f( x ) dx$ is
$2$
$3$
$4$
more than $4$
The value of integral $\int_0^1 {{e^{{x^2}}}} dx$ lies in interval
Let $\frac{d}{{dx}}F(x) = \left( {\frac{{{e^{\sin x}}}}{x}} \right)\,;\,x > 0$. If $\int_{\,1}^{\,4} {\frac{3}{x}{e^{\sin {x^3}}}dx = F(k) - F(1)} $, then one of the possible value of $k$, is
Let $f: R \rightarrow R$ be a function defined as $f(x)=a \sin \left(\frac{\pi[x]}{2}\right)+[2-x], a \in R$, where [t] is the greatest integer less than or equal to $t$. If $\lim _{x \rightarrow-1} f(x)$ exists, then the value of $\int_{0}^{4} f(x) d x$ is equal to.
Let $f$ be a continuous function defined on $[0,1]$ such that $\int_0^1 f^2(x) d x=\left(\int_0^1 f(x) d x\right)^2$. Then, the range of $f$