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Step-by-Step Solution
Step 1: Express the integrals
We are given two definite integrals:
$I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}}\,dx \quad \text{and} \quad J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}}\,dx.$
Step 2: Compare $\frac{\sin x}{\sqrt{x}}$ with $\sqrt{x}$
For $x \in (0,1)$, we know $\sin x < x$. Thus:
$\displaystyle \frac{\sin x}{\sqrt{x}} < \frac{x}{\sqrt{x}} = \sqrt{x}.$
Step 3: Integrate the upper bound for $I$
Using the inequality from Step 2, we have:
$\displaystyle I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}}\,dx < \int_{0}^{1} \sqrt{x}\,dx.$
Now, compute $\int_{0}^{1} \sqrt{x}\,dx$:
$\displaystyle \int_{0}^{1} \sqrt{x}\,dx = \int_{0}^{1} x^{1/2}\,dx
= \left[\frac{2\,x^{3/2}}{3}\right]_{0}^{1}
= \frac{2}{3}.$
Hence, $I < \frac{2}{3}.$
Step 4: Compare $\frac{\cos x}{\sqrt{x}}$ with $\frac{1}{\sqrt{x}}$
Similarly, for $x \in (0,1)$, we have $\cos x < 1$. Thus:
$\displaystyle \frac{\cos x}{\sqrt{x}} < \frac{1}{\sqrt{x}}.$
Step 5: Integrate the upper bound for $J$
Using the above inequality:
$\displaystyle J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}}\,dx < \int_{0}^{1} \frac{1}{\sqrt{x}}\,dx.$
Now, compute $\int_{0}^{1} x^{-1/2}\,dx$:
$\displaystyle \int_{0}^{1} x^{-1/2}\,dx
= \left[2\,\sqrt{x}\right]_{0}^{1}
= 2.$
Hence, $J < 2.$
Step 6: Conclude the values of $I$ and $J$
From the above steps,
$I < \frac{2}{3}.$
$J < 2.$
Therefore, the correct choice is:
$I < \frac{2}{3} \quad \text{and} \quad J < 2.$
