1. The given integral is:
\( I(R) = \int_{0}^{R} e^{-R \sin x} \, dx. \)
2. The term \( e^{-R \sin x} \) involves an exponential function with an oscillating argument \( \sin x \). The oscillatory nature of \( \sin x \) leads to variable behavior of the integrand \( e^{-R \sin x} \), which complicates direct evaluation.
3. To evaluate this integral analytically:
4. The expression \( \frac{\pi}{2R}(1 - e^{-R}) \) comes from approximations often used for integrals with oscillatory terms, but it is not exact.
5. Since \( I(R) \) and \( \frac{\pi}{2R}(1 - e^{-R}) \) involve different behaviors depending on \( R \), they cannot be directly compared for all values of \( R > 0 \).
For the beam and loading shown in the figure, the second derivative of the deflection curve of the beam at the mid-point of AC is given by \( \frac{\alpha M_0}{8EI} \). The value of \( \alpha \) is ........ (rounded off to the nearest integer).
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to:
The integral is given by:
\[ 80 \int_{0}^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16 \sin 2\theta} d\theta \]
is equals to?
\( \text{M} \xrightarrow{\text{CH}_3\text{MgBr}} \text{N} + \text{CH}_4 \uparrow \xrightarrow{\text{H}^+} \text{CH}_3\text{COCH}_2\text{COCH}_3 \)
Identify the ion having 4f\(^6\) electronic configuration.