Key Concepts and Formulas

• Product Rule for Differentiation: $\frac{d}{dt}(u(t)v(t)) = u'(t)v(t) + u(t)v'(t)$
• Fundamental Theorem of Calculus (Part 2): $\int_a^b h'(t)\,dt = h(b) - h(a)$
• L'Hopital's Rule: For limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, $\lim_{x \to c} \frac{F(x)}{H(x)} = \lim_{x \to c} \frac{F'(x)}{H'(x)}$ (if the latter limit exists).
• Trigonometric Identities and Values: $\sec x = \frac{1}{\cos x}$, $\sin(\pi/2) = 1$, $\cos(\pi/2) = 0$, $\sec(\pi/4) = \sqrt{2}$.

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Step-by-Step Solution

1. Simplify the Integrand

We are given the integral for $g(x)$: $g(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt}$ Let's examine the integrand: $f'(t)\sec t + \tan t\sec t\,f(t)$. We can recognize this as the result of the product rule for differentiation. Consider the function $h(t) = f(t)\sec t$. Using the product rule, $h'(t) = \frac{d}{dt}(f(t)\sec t) = f'(t)\sec t + f(t) \cdot (\sec t \tan t)$. This precisely matches the integrand. Therefore, the integrand is the derivative of $f(t)\sec t$.

Why this step is taken: Recognizing the integrand as a derivative of a product simplifies the integration process immensely, allowing us to use the Fundamental Theorem of Calculus directly.

2. Evaluate the Definite Integral

Now, we can apply the Fundamental Theorem of Calculus to evaluate $g(x)$: $g(x) = \left[f(t)\sec t\right]_x^{\pi/4}$ $g(x) = f\left(\frac{\pi}{4}\right)\sec\left(\frac{\pi}{4}\right) - f(x)\sec(x)$ We are given $f(\pi/4) = \sqrt{2}$ and we know $\sec(\pi/4) = \sqrt{2}$. Substituting these values: $g(x) = (\sqrt{2})(\sqrt{2}) - f(x)\sec(x)$ $g(x) = 2 - f(x)\sec(x)$

Why this step is taken: This step expresses $g(x)$ in a much simpler form, which is crucial for the subsequent limit evaluation.

3. Set up the Limit and Identify the Indeterminate Form

We need to find $\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)$. Substituting the simplified expression for $g(x)$: $\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x) = \mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} \left(2 - f(x)\sec(x)\right)$ As $x \to (\pi/2)^-$:

• $f(x) \to f(\pi/2) = 0$ (given).
• $\sec(x) = \frac{1}{\cos x}$. As $x \to (\pi/2)^-$, $\cos x \to 0^+$, so $\sec x \to +\infty$.

The term $f(x)\sec(x)$ approaches the indeterminate form $0 \cdot \infty$. To evaluate this limit, we rewrite it as a fraction to apply L'Hopital's Rule. $f(x)\sec(x) = \frac{f(x)}{\cos x}$ As $x \to (\pi/2)^-$, this fraction takes the indeterminate form $\frac{0}{0}$.

Why this step is taken: Identifying the indeterminate form is essential for choosing the correct method to evaluate the limit. Rewriting $f(x)\sec(x)$ as $\frac{f(x)}{\cos x}$ prepares it for L'Hopital's Rule.

4. Apply L'Hopital's Rule

Let $L = \mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} \frac{f(x)}{\cos x}$. Since this is a $\frac{0}{0}$ indeterminate form, we apply L'Hopital's Rule: $L = \mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} \frac{\frac{d}{dx}(f(x))}{\frac{d}{dx}(\cos x)}$ $L = \mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} \frac{f'(x)}{-\sin x}$ Now, we substitute the given values and known trigonometric values as $x \to (\pi/2)^-$: $f'(x) \to f'(\pi/2) = 1$ (given). $-\sin x \to -\sin(\pi/2) = -1$. So, $L = \frac{1}{-1} = -1$

Why this step is taken: L'Hopital's Rule provides a direct method to evaluate the limit of the indeterminate form by taking the derivatives of the numerator and denominator.

5. Calculate the Final Limit

Now we substitute the value of $L$ back into the limit expression for $g(x)$: $\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x) = 2 - L$ $\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x) = 2 - (-1)$ $\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x) = 2 + 1 = 3$

Why this step is taken: This final step combines the constant term from the simplified $g(x)$ with the evaluated limit of the indeterminate part to arrive at the final answer.

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Common Mistakes & Tips

• Incorrectly identifying the integrand: Failing to recognize the integrand as a derivative of a product will make the integration significantly harder, or impossible with elementary functions.
• Errors in L'Hopital's Rule: Forgetting to differentiate both the numerator and the denominator, or making sign errors during differentiation (e.g., derivative of $\cos x$ is $-\sin x$), can lead to incorrect limit values.
• Ignoring indeterminate forms: Directly substituting values into expressions that yield indeterminate forms like $0 \cdot \infty$ will not yield the correct limit. Always convert to $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

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Summary

The problem involves evaluating a definite integral where the integrand is recognized as the derivative of a product, $f(t)\sec t$. After applying the Fundamental Theorem of Calculus, the function $g(x)$ simplifies to $2 - f(x)\sec x$. The limit of $g(x)$ as $x$ approaches $\pi/2$ from the left involves an indeterminate form $0 \cdot \infty$, which is transformed into $\frac{0}{0}$ and resolved using L'Hopital's Rule. The final limit is found to be 3.

The final answer is $\boxed{3}$.