The area bounded by the graph of `y=f(x), f(x) gt0` on [0,a] and x-axis is `(a^(2))/(2)+(a)/(2) sin a +(pi)/(2) cos a ` then find the value of `f((pi)/(2))`.

Explanation

To find the value of \( f\left(\frac{\pi}{2}\right) \), we start with the given area bounded by the graph of \( y = f(x) \) on the interval \([0, a]\) and the x-axis, which is expressed as:

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

### Step 1: Set up the equation for the area
The area under the curve from \( 0 \) to \( a \) can be expressed as:

\[ \int_0^a f(x) \, dx = \frac{a^2}{2} + \frac{a}{2} \sin a + \frac{\pi}{2} \cos a \]

### Step 2: Differentiate both sides with respect to \( a \)
To find \( f(a) \), we differentiate both sides of the equation with respect to \( a \):

\[ \frac{d}{da} \left( \int_0^a f(x) \, dx \right) = f(a) \]

Using the Fundamental Theorem of Calculus, we have:

\[ f(a) = \frac{d}{da} \left( \frac{a^2}{2} + \frac{a}{2} \sin a + \frac{\pi}{2} \cos a \right) \]

### Step 3: Differentiate the right-hand side
Now we differentiate the right-hand side term by term:

1. The derivative of \( \frac{a^2}{2} \) is \( a \).
2. For \( \frac{a}{2} \sin a \), we apply the product rule:
- Let \( u = \frac{a}{2} \) and \( v = \sin a \).
- Then \( \frac{du}{da} = \frac{1}{2} \) and \( \frac{dv}{da} = \cos a \).
- So, \( \frac{d}{da}(uv) = u \frac{dv}{da} + v \frac{du}{da} = \frac{a}{2} \cos a + \sin a \cdot \frac{1}{2} \).

3. The derivative of \( \frac{\pi}{2} \cos a \) is \( -\frac{\pi}{2} \sin a \).

Combining these, we have:

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

### Step 4: Simplify the expression for \( f(a) \)
Now, we can simplify \( f(a) \):

\[ f(a) = a + \frac{a}{2} \cos a + \frac{1}{2} \sin a - \frac{\pi}{2} \sin a \]

Combining the sine terms gives:

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

### Step 5: Substitute \( a = \frac{\pi}{2} \)
Now we substitute \( a = \frac{\pi}{2} \):

\[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + \frac{\frac{\pi}{2}}{2} \cos\left(\frac{\pi}{2}\right) + \left( \frac{1}{2} - \frac{\pi}{2} \right) \sin\left(\frac{\pi}{2}\right) \]

Calculating each term:

1. \( \cos\left(\frac{\pi}{2}\right) = 0 \)
2. \( \sin\left(\frac{\pi}{2}\right) = 1 \)

Thus, we have:

\[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + 0 + \left( \frac{1}{2} - \frac{\pi}{2} \right) \cdot 1 \]

This simplifies to:

\[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + \frac{1}{2} - \frac{\pi}{2} = \frac{1}{2} \]

### Final Answer
Thus, the value of \( f\left(\frac{\pi}{2}\right) \) is:

\[ \boxed{\frac{1}{2}} \]