Problem 2 - Calculus

Question

Find the derivatives of the following functions \(F(x)\). $$ \int_{x}^{1} \cos 3 t d t $$

Step-by-Step Solution

Answer

The derivative is \( F'(x) = -3\cos(3x) \).

Step 1: Understanding the problem

We need to find the derivative of a function defined as an integral with a variable limit of integration. The function given is \( F(x) = \int_{x}^{1} \cos(3t) \, dt \), and we are to find \( F'(x) \).

Step 2: Applying the Fundamental Theorem of Calculus Part 1

According to the Fundamental Theorem of Calculus Part 1, for a function \( G(x) = \int_{a}^{x} f(t) \, dt \), the derivative is \( G'(x) = f(x) \). However, our function has a variable lower limit, \( x \), and a constant upper limit, \( 1 \).

Step 3: Adjusting limits of integration

To use the theorem, we adjust the limits to fit the form \( \int_{a}^{x} \). This is done by negating the integral: \( F(x) = -\int_{1}^{x} \cos(3t) \, dt \). Then the derivative is \( F'(x) = -\cos(3x) \).

Step 4: Applying the Chain Rule

Since the integrand is \( \cos(3x) \), we also need to apply the chain rule as the inside function is \( 3x \). Thus, \( F'(x) = -\cos(3x) \times 3 = -3 \cos(3x) \).

Key Concepts

Derivatives of Integrals

Understanding derivatives of integrals involves applying the Fundamental Theorem of Calculus. In essence, when you have a function defined by an integral, you can find its derivative in a particular way.
For example, say you have a function like \( G(x) = \int_{a}^{x} f(t) \, dt \). According to the theorem, the derivative of this function \( G'(x) \) is simply \( f(x) \). So, the derivative is the integrand \( f(t) \) evaluated at \( x \).
This relationship gives us a straightforward tool to differentiate functions expressed as integrals when the limits are from a constant to a variable. But this becomes slightly more intricate when the expression involves variable limits, which we will detail in the next section.
Remember:

• To find \( G'(x) \), just replace the integrand's variable with \( x \).
• This approach only works directly when the upper limit of the integral is a variable, and the lower is constant.

Variable Limits of Integration

When dealing with integrals where limits of integration are variable, it requires slight adjustment to the standard approach. Consider the function \( F(x) = \int_{x}^{1} \cos(3t) \, dt \). Here, the limit \( x \) is at the lower end.
To apply the Fundamental Theorem of Calculus, the integral should ideally be from a constant to a variable upper limit. This situation is handled by reversing the limits, which introduces a negative sign. That converts our example to \( F(x) = -\int_{1}^{x} \cos(3t) \, dt \).
In reversing the limits, notice:

• The overall effect is introducing a negative sign before the integral.
• The order of integration matters; it switches from the start to the end at \( x \).
• Once limits are rearranged, the Fundamental Theorem applies smoothly.

This allows us to differentiate the integral straightforwardly with respect to its variable limit, preparing the way for applying further differentiation rules like the chain rule.

Chain Rule in Calculus

After adjusting the limits of integration, it's often critical to use the chain rule to find the correct derivative. In functions like \( F(x) = -\int_{1}^{x} \cos(3t) \, dt \), you encounter a composite function because the integrand involves another function of \( x \).
Here, the integrand is \( \cos(3x) \). For differentiation, you follow these steps:

• Find the derivative directly: \(-\cos(3x) \)
• Apply the chain rule: Multiply by the derivative of the inside function \( 3x \), which is \( 3 \).
• Thus, the overall derivative becomes \(-3 \cos(3x) \).

The chain rule simply says that if you differentiate a composite function, you must also consider the derivative of the inner function.
This step ensures that the final derivative is completely evaluated concerning the original variable \( x \), capturing the integral's changing nature and the function's depth.