Problem 30 - Calculus Single Variable

Question

In each of Exercises \(29-34,\) calculate the derivative of \(F(x)\) with respect to \(x\) $$ F(x)=\int_{x}^{\pi / 4} \sqrt{1+\cos (t)} d t $$

Step-by-Step Solution

Answer

\( F'(x) = -\sqrt{1+\cos(x)} \).

Step 1: Understand the Problem

We need to differentiate a function defined by an integral with a variable lower limit using the Fundamental Theorem of Calculus Part 1. The function given is \( F(x) = \int_{x}^{\pi/4} \sqrt{1+\cos(t)} \, dt \).

Step 2: Apply the Fundamental Theorem of Calculus Part 1

The Fundamental Theorem of Calculus Part 1 states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). However, since the variable limit is the lower limit in our integral, we'll use \( -f(x) \) after applying the theorem.

Step 3: Identify the Integrand Function

Identify \( f(t) \), which is the integrand function. In the given integral, \( f(t) = \sqrt{1+\cos(t)} \).

Step 4: Differentiate with Respect to Variable x

Since the integrand \( f(t) \) is \( \sqrt{1+\cos(t)} \) and it's integrated from \( x \) to a constant, the derivative with respect to \( x \) is \( -\sqrt{1+\cos(x)} \).

Step 5: Write Down the Final Derivative

Thus, the derivative \( F'(x) \) is given by \( -\sqrt{1+\cos(x)} \).

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus serves as a vital link between differentiation and integration, two core concepts of calculus. It is usually presented in two parts.
Part 1 of this theorem, which is applicable in our problem, states that if you have a function defined as an integral with a variable upper limit, like this:

• \( F(x) = \int_{a}^{x} f(t) \, dt \)

Then, the derivative of this function with respect to \( x \) is simply the integrand evaluated at \( x \), that is:

• \( F'(x) = f(x) \)

In our specific example, the theorem is applied in a slightly modified manner because the variable \( x \) is in the lower limit. This is where the theorem indicates a negative sign: the derivative \( F'(x) \) becomes \(-f(x)\).
This part of the theorem tells us how easy it can be to differentiate an integral by merely substituting the upper variable with the integrand.

Variable Limits of Integration

Integrals with variable limits, especially where the variable acts as a limit of integration, can be intriguing. Imagine your function has bounds like in our original exercise:

• \( F(x) = \int_{x}^{\pi/4} \sqrt{1+\cos(t)} \, dt \)

Here, the variable \( x \) serves as the lower limit of the integration. When working with variable limits, the integration bounds are not constants. Instead, they move or expand as the variable changes. This characteristic affects the outcome when you differentiate these integrals.
Because the derivative of an integral with a variable boundary changes based on whether the variable is at the upper or lower limit, you must adjust the resulting derivative. For our problem, having \( x \) as the lower limit requires us to apply the rule from the Fundamental Theorem of Calculus, using that negative sign as part of the resultant derivative.

Differentiating Under the Integral Sign

Differentiating under the integral sign is another technique that offers flexibility in calculus. It's particularly useful in scenarios as seen in the given exercise, where the integral takes on variable limits. The problem with \( F(x) = \int_{x}^{\pi/4} \sqrt{1+\cos(t)} \, dt \) showcases this process.When you differentiate an integral where one of the limits is a variable, you essentially change the dynamic of the function and its derivative behavior shifts accordingly. In this exercise, since \( x \) is the lower bound, applying the differentiation in this context means reversing the sign of the integrand function after applying the fundamental theorem, resulting in \(-\sqrt{1+\cos(x)}\). This approach shines when dealing with complex integrals, making the derivative straightforward to compute once you understand the structure of the integral itself.