Key Concepts and Formulas

• Fundamental Theorem of Calculus (Part 1): If $F(x) = \int_a^x f(t) dt$, then $F'(x) = f(x)$.
• Leibniz Rule (a special case): If $G(x) = \int_a^x f(t) dt$, then $G'(x) = f(x)$.
• Differentiation: The derivative of a product $uv$ is $u'v + uv'$.

Step-by-Step Solution

Step 1: Define the Area Function

We are given that the area bounded by the curve $y = f(x)$, the x-axis, and the ordinates $x = \frac{\pi}{4}$ and $x = \beta > \frac{\pi}{4}$ is given by: $A(\beta) = \int_{\pi/4}^{\beta} f(x) dx = \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta$ Our goal is to find the value of $f\left(\frac{\pi}{2}\right)$.

Step 2: Differentiate both sides with respect to $\beta$

We will use the Fundamental Theorem of Calculus (Part 1) or Leibniz rule (a special case where the lower bound is constant) on the left-hand side and standard differentiation on the right-hand side. Differentiating both sides of the equation with respect to $\beta$, we get: $\frac{d}{d\beta} \left[ \int_{\pi/4}^{\beta} f(x) dx \right] = \frac{d}{d\beta} \left[ \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \right]$ Using the Fundamental Theorem of Calculus (or Leibniz Rule), the derivative of the integral on the left side is simply $f(\beta)$. On the right side, we differentiate term by term using the product rule for $\beta \sin \beta$: $f(\beta) = \frac{d}{d\beta} (\beta \sin \beta) + \frac{d}{d\beta} \left(\frac{\pi}{4} \cos \beta\right) + \frac{d}{d\beta} (\sqrt{2} \beta)$ $f(\beta) = (\sin \beta + \beta \cos \beta) + \left(-\frac{\pi}{4} \sin \beta\right) + \sqrt{2}$ Therefore, $f(\beta) = \sin \beta + \beta \cos \beta - \frac{\pi}{4} \sin \beta + \sqrt{2}$

Step 3: Substitute $\beta = \frac{\pi}{2}$

To find $f\left(\frac{\pi}{2}\right)$, we substitute $\beta = \frac{\pi}{2}$ into the expression for $f(\beta)$: $f\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) - \frac{\pi}{4} \sin\left(\frac{\pi}{2}\right) + \sqrt{2}$ Since $\sin\left(\frac{\pi}{2}\right) = 1$ and $\cos\left(\frac{\pi}{2}\right) = 0$, we have: $f\left(\frac{\pi}{2}\right) = 1 + \frac{\pi}{2}(0) - \frac{\pi}{4}(1) + \sqrt{2}$ $f\left(\frac{\pi}{2}\right) = 1 - \frac{\pi}{4} + \sqrt{2}$ $f\left(\frac{\pi}{2}\right) = \left(1 - \frac{\pi}{4} + \sqrt{2}\right)$ $f\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{4} + \sqrt{2} - 1\right)$

Common Mistakes & Tips

• Product Rule: Don't forget to apply the product rule when differentiating terms like $\beta \sin \beta$. A common mistake is to only differentiate one part of the product.
• Sign Errors: Pay close attention to signs when differentiating trigonometric functions. The derivative of $\cos x$ is $-\sin x$.
• Leibniz Rule: Recognize when to apply Leibniz's rule (or, more simply, the Fundamental Theorem of Calculus Part 1) when differentiating an integral with a variable limit.

Summary

We were given an expression for the area under a curve as a function of the upper limit of integration. To find the value of the function at a specific point, we differentiated the area function with respect to the upper limit using the Fundamental Theorem of Calculus (or Leibniz Rule). This allowed us to isolate the function $f(x)$ and then substitute the desired value, $\frac{\pi}{2}$, to obtain the final answer. The result is $f\left(\frac{\pi}{2}\right) = \left(1 - \frac{\pi}{4} + \sqrt{2}\right)$.

Final Answer

The final answer is \boxed{\left(1 - {\pi \over 4} + \sqrt 2 \right)}, which corresponds to option (D).