Animated Solution for Mathematics - Definite Integration: Let $F(x)={\int }_x^{x^2+\frac{\pi }{6}}2{\cos }^{2}t(dt)$ for all $x\in R$ and $f:[0,1/2]\to [0,\infty )$ be a continuous function. For $\alpha \in [0,1/2]$, if $F^{\prime }(\alpha )+2$ is the area of the region bounded by $x=0,y=0,y=f(x)$ and $x=\alpha$, then $f(0)$ is

Welcome, fellow explorer of the mathematical universe. Today, we are going to unravel a problem that might look like a tangled mess of integrals and derivatives, but beneath the surface, it is a beautifully choreographed dance between geometry and analysis.

We are tasked with finding $f(0)$, given a complex integral definition for $F(x)$ and a relationship involving the area under $f(x)$. Let us break this down, step by step, and see how the pieces fall into place.

Imagine a curve representing the function $f(x)$. We are interested in the area bounded by this curve, the $x$-axis, the $y$-axis, and a vertical line at $x=\alpha$.

Mathematically, we define this area as:

The problem provides the relationship $A(\alpha )=F^{\prime }(\alpha )+2$. To find $f(0)$, we need to connect $f$ to $F$.

If we differentiate both sides of this equation with respect to $\alpha$, the Fundamental Theorem of Calculus tells us that the derivative of the integral is simply $f(\alpha )$. On the right side, the derivative of $F^{\prime }(\alpha )+2$ is $F^{\prime \prime }(\alpha )$ (since the derivative of the constant $2$ is zero).

Thus, we arrive at our target: To find $f(0)$, we simply need to calculate $F^{\prime \prime }(0)$.

Now, we turn our attention to the definition of $F(x)$:

This is where we invoke the Leibniz Rule for differentiating under the integral sign. This rule is our secret weapon:

Here, our integrand $w(t)$ is $2{\cos }^{2}t$, our lower limit $g(x)$ is $x$, and our upper limit $h(x)$ is $x^2+\frac{\pi }{6}$.

Let us apply the rule. First, we evaluate the integrand at the upper limit and multiply by the derivative of the upper limit:

Next, we subtract the integrand evaluated at the lower limit, multiplied by the derivative of the lower limit:

Putting it together, we get:

We need $F^{\prime \prime }(x)$, so we differentiate $F^{\prime }(x)$ again. Let us split $F^{\prime }(x)$ into two terms: $T_1=4x{\cos }^2(x^2+\frac{\pi }{6})$ and $T_2=-2{\cos }^{2}x$.

For $T_1$, we use the product rule:

For $T_2$, we use the chain rule:

Now, we combine them:

Finally, we evaluate at $x=0$. The term with $4x$ vanishes, and $2\sin (2\cdot 0)$ is also zero. We are left with:

Since $\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}$, then ${\cos }^2(\frac{\pi }{6})=\frac{3}{4}$. Thus:

The complexity dissolves, leaving us with a clean, elegant integer. The final answer is 3.