We are asked to find the area of the region bounded by the curve \( y = 5x \), the x-axis, and \( x = 4 \).
The area under the curve is given by the integral:
\( \text{Area} = \int_0^4 5x \, dx \).
Now, compute the integral:
\( \int 5x \, dx = \frac{5x^2}{2} \).
Evaluate the integral from 0 to 4:
\( \left[ \frac{5x^2}{2} \right]_0^4 = \frac{5(4^2)}{2} - \frac{5(0^2)}{2} = \frac{5(16)}{2} = 40 \).
The correct answer is 40.
In the adjoining figure, TP and TQ are tangents drawn to a circle with centre O. If $\angle OPQ = 15^\circ$ and $\angle PTQ = \theta$, then find the value of $\sin 2\theta$. 
What is the angle between the hour and minute hands at 4:30?