Answer
$\dfrac{1}{2}$
Work Step by Step
When the function $f(x)$ has minimum and maximum on $[a,b]$ then , we have
$min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$
Given: $f(x)=\sin x \leq x$ for $x \geq 0$
or, $\sin x-x \leq 0$ for $x \geq 0$
Now, we have
$ \int_0^1 (\sin x-x) dx \leq 0$
or, $\int_0^1 \sin x dx -\int_0^1 x dx \leq 0$
or, $\int_0^1 \sin x dx \leq \int_0^1 x dx$
Thus, $\int_0^1 \sin x dx \leq \dfrac{1}{2}$
Hence, the upper bound is: $\dfrac{1}{2}$
