Question

Let \( f(x) \) be a non-negative continuous function of real variable \( x \). If the area under the curve \( y = f(x) \) from \( x = 0 \) to \( x = a \) is \( \frac{a^2}{2} + \frac{a}{2} \sin a + \frac{\pi}{2} \cos a - \frac{\pi}{2} \), then the value of \( f \left( \frac{\pi}{2} \right) \) is \(\underline{\hspace{2cm}}\) (round off to one decimal place).

Hint:

To find the value of a function at a point, differentiate the area function with respect to the upper limit and substitute the point into the result.

Answer 1

Correct Answer: 0.5

We are given the area under the curve as:
\[ \int_0^a f(x) \, dx = \frac{a^2}{2} + \frac{a}{2} \sin a + \frac{\pi}{2} \cos a - \frac{\pi}{2} \] We need to find \( f \left( \frac{\pi}{2} \right) \). To do this, we differentiate the area expression with respect to \( a \): \[ f(a) = \frac{d}{da} \left( \frac{a^2}{2} + \frac{a}{2} \sin a + \frac{\pi}{2} \cos a - \frac{\pi}{2} \right) \] After differentiating:
\[ f(a) = a + \frac{1}{2} \sin a + \frac{a}{2} \cos a \] Now, substitute \( a = \frac{\pi}{2} \): \[ f\left( \frac{\pi}{2} \right) = \frac{\pi}{2} + \frac{1}{2} \sin \frac{\pi}{2} + \frac{\pi}{2} \cos \frac{\pi}{2} = \frac{\pi}{2} + \frac{1}{2} = 0.5 \] Thus, the value of \( f\left( \frac{\pi}{2} \right) \) is \( 0.5 \).