Let f (x) be a continous function such that the area bounded by the curve `y =f (x)` x-axis and the lines x=0 and `x =a is (a ^(2))/( 2) + (a)/(2) sin a+ pi/2 cos a,` then `f ((pi)/(2))` is

Answer

To solve the problem, we need to find the value of \( f\left(\frac{\pi}{2}\right) \) given that the area bounded by the curve \( y = f(x) \), the x-axis, and the lines \( x = 0 \) and \( x = a \) is given by:

\[ \text{Area} = \frac{a^2}{2} + \frac{a}{2} \sin a + \frac{\pi}{2} \cos a \]

### Step-by-Step Solution:

1. **Understand the Area Representation**:
The area under the curve from \( x = 0 \) to \( x = a \) can be expressed as:
\[ \int_0^a f(x) \, dx = \frac{a^2}{2} + \frac{a}{2} \sin a + \frac{\pi}{2} \cos a \]

2. **Differentiate Both Sides**:
To find \( f(a) \), we differentiate both sides with respect to \( a \):
\[ \frac{d}{da} \left( \int_0^a f(x) \, dx \right) = \frac{d}{da} \left( \frac{a^2}{2} + \frac{a}{2} \sin a + \frac{\pi}{2} \cos a \right) \]

By the Fundamental Theorem of Calculus, the left-hand side becomes:
\[ f(a) \]

3. **Differentiate the Right-Hand Side**:
Now we differentiate the right-hand side term by term:
- The derivative of \( \frac{a^2}{2} \) is \( a \).
- For \( \frac{a}{2} \sin a \), we use the product rule:
\[ \frac{d}{da} \left( \frac{a}{2} \sin a \right) = \frac{1}{2} \sin a + \frac{a}{2} \cos a \]
- The derivative of \( \frac{\pi}{2} \cos a \) is:
\[ -\frac{\pi}{2} \sin a \]

Putting it all together, we have:
\[ f(a) = a + \left( \frac{1}{2} \sin a + \frac{a}{2} \cos a \right) - \frac{\pi}{2} \sin a \]

Simplifying this gives:
\[ f(a) = a + \frac{1}{2} \sin a + \frac{a}{2} \cos a - \frac{\pi}{2} \sin a \]
\[ f(a) = a + \left( \frac{1}{2} - \frac{\pi}{2} \right) \sin a + \frac{a}{2} \cos a \]

4. **Substitute \( a = \frac{\pi}{2} \)**:
Now we find \( f\left(\frac{\pi}{2}\right) \):
\[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + \left( \frac{1}{2} - \frac{\pi}{2} \right) \sin\left(\frac{\pi}{2}\right) + \frac{\frac{\pi}{2}}{2} \cos\left(\frac{\pi}{2}\right) \]

We know:
- \( \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)

Thus, substituting these values gives:
\[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + \left( \frac{1}{2} - \frac{\pi}{2} \right) \cdot 1 + 0 \]
\[ = \frac{\pi}{2} + \frac{1}{2} - \frac{\pi}{2} \]
\[ = \frac{1}{2} \]

### Final Answer:
\[ f\left(\frac{\pi}{2}\right) = \frac{1}{2} \]