Q.56

Problem Statement

Evaluate the integral
\[
\int_{1}^{\sqrt{3}} \frac{1 + [x]}{1 + x^{2}}\,dx = L,
\]
where \([x]\) denotes the greatest integer less than or equal to \(x\). Then express \(L\) in degrees.

Step 1: Understand the Greatest Integer Function

On the interval \([1,\sqrt{3}]\), the variable \(x\) satisfies \(1 \le x < 2\), because \(\sqrt{3} \approx 1.732 < 2\).
For every \(x\) in this interval, the greatest integer less than or equal to \(x\) is \([x] = 1\).
Therefore the numerator of the integrand simplifies as \(1 + [x] = 1 + 1 = 2\).
So the integral becomes
\[
L = \int_{1}^{\sqrt{3}} \frac{2}{1 + x^{2}}\,dx.
\]

Step 2: Integrate the Simplified Function

The antiderivative of \(\dfrac{1}{1+x^{2}}\) is \(\tan^{-1}x\). Thus,
\[
\int \frac{2}{1 + x^{2}}\,dx = 2\tan^{-1}x + C.
\]
Apply the limits from \(1\) to \(\sqrt{3}\):
\[
L = \left[2\tan^{-1}x\right]_{1}^{\sqrt{3}}
= 2\left(\tan^{-1}(\sqrt{3}) – \tan^{-1}(1)\right).
\]
Using standard values,
\(\tan^{-1}(1) = \dfrac{\pi}{4}\) and \(\tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}\). Therefore,
\[
L = 2\left(\frac{\pi}{3} – \frac{\pi}{4}\right)
= 2\left(\frac{4\pi – 3\pi}{12}\right)
= 2\cdot\frac{\pi}{12}
= \frac{\pi}{6} \text{ radians}.
\]

Step 3: Convert L from Radians to Degrees

To convert from radians to degrees, use
\[
\text{Degrees} = \text{Radians} \times \frac{180^\circ}{\pi}.
\]
Here \(L = \dfrac{\pi}{6}\) radians, so
\[
L = \frac{\pi}{6} \times \frac{180^\circ}{\pi}
= \frac{180^\circ}{6}
= 30^\circ.
\]
Alternatively, in degrees directly:
\(\tan^{-1}(\sqrt{3}) = 60^\circ\) and \(\tan^{-1}(1) = 45^\circ\). Hence
\[
L = 2(60^\circ – 45^\circ)
= 2 \times 15^\circ
= 30^\circ.
\]

Final Answer

\[
\boxed{L = 30^\circ}
\]