Let I = _0^1 x x,dx and J = _0^1 x x,dx.

Question

Let $I = \displaystyle\int_0^1 \dfrac{\sin x}{\sqrt x}\,dx$ and $J = \displaystyle\int_0^1 \dfrac{\cos x}{\sqrt x}\,dx$. Then which one of the following is true?

Accepted Answer

We know that $${{\sin x} \over x} < 1,$$ for $$x \in \left( {0,1} \right)$$

$$ \Rightarrow {{\sin x} \over {\sqrt x }} < \sqrt x $$ on $$x \in \left( {0,1} \right)$$

$$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx < \int\limits_0^1 {\sqrt x dx}  = \left[ {{{2{x^{3/2}}} \over 3}} \right]} _0^1$$

$$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}} dx < {2 \over 3} \Rightarrow I < {2 \over 3}$$

Also $${{\cos x} \over {\sqrt x }} < {1 \over {\sqrt x }}$$ for $$x \in \left( {0,1} \right)$$

$$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < \int\limits_0^1 {{x^{ - 1/2}}dx} } $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {2\sqrt x } \right]_0^1 = 2$$

$$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < 2} $$

$$ \Rightarrow J < 2$$

Solutions · 1