Finding \(f(\frac{\pi}{2})\) from Area Under Curve

The area under the curve $y = f(x)$ from $x = 0$ to $x = a$ is given by the definite integral $A(a) = \int_0^a f(x) dx$. According to the Fundamental Theorem of Calculus, the derivative of the area function $A(a)$ with respect to $a$ gives the function value $f(a)$, i.e., $f(a) = \frac{dA}{da}$.

Area Function Definition

The given area function is:

$ A(a) = \frac{a^2}{\pi} + \frac{a}{\pi} \sin a + \frac{2}{\pi} \cos a - \frac{2}{\pi} $

Calculating the Function \(f(a)\)

We need to differentiate $A(a)$ with respect to $a$ to find $f(a)$.

• Derivative of $\frac{a^2}{\pi}$ is $\frac{2a}{\pi}$.

• Derivative of $\frac{a}{\pi} \sin a$ using the product rule is $\frac{1}{\pi}(\sin a + a \cos a)$.

• Derivative of $\frac{2}{\pi} \cos a$ is $-\frac{2}{\pi} \sin a$.

• Derivative of the constant $-\frac{2}{\pi}$ is $0$.

Combining these, we get:

$ f(a) = \frac{dA}{da} = \frac{2a}{\pi} + \frac{1}{\pi}(\sin a + a \cos a) - \frac{2}{\pi} \sin a $

Simplifying the expression:

$ f(a) = \frac{2a}{\pi} + \frac{a \cos a}{\pi} + \frac{\sin a}{\pi} - \frac{2 \sin a}{\pi} $

$ f(a) = \frac{2a}{\pi} + \frac{a \cos a}{\pi} - \frac{\sin a}{\pi} $

Evaluating \(f(\frac{\pi}{2})\)

Now substitute $a = \frac{\pi}{2}$ into the expression for $f(a)$:

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

Since $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$:

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

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

$ f\left(\frac{\pi}{2}\right) = 1 - \frac{1}{\pi} $

Final Result and Rounding

Calculate the numerical value:

$ 1 - \frac{1}{\pi} \approx 1 - \frac{1}{3.14159} \approx 1 - 0.3183 \approx 0.6817 $

Rounding the result to one decimal place:

$ f\left(\frac{\pi}{2}\right) \approx 0.7 $