Problem 6 - Active Calculus

Question

Use the Fundamental Theorem of Calculus to find \(\int_{5 \pi / 2}^{3 \pi} e^{\sin (q)} \cdot \cos (q) d q=\) _____________________

Step-by-Step Solution

Answer

The integral evaluates to \(1 - e\).

Step 1 - Identify the Integrand

The function to be integrated is \(e^{\sin(q)} \cdot \cos(q)\). This is the integrand.

Step 2 - Recognize the Inner Function

Notice that \(\sin(q)\) is the inner function inside the exponent and \(\cos(q)\) is its derivative.

Step 3 - Set Up for Substitution

Let \(u = \sin(q)\). Then, \(du = \cos(q) dq\). This changes the integral to \int e^{u} du\.

Step 4 - Change the Limits of Integration

Convert the limits of integration from \(q\) to \(u\). When \( q = 5 \pi / 2\), \sin(5 \pi / 2) = 1\. When \( q = 3\pi \), \sin(3\pi) = 0\. Thus, the new limits of integration are from 1 to 0.

Step 5 - Integrate the New Function

Integrate \int e^{u} du\ to get \( e^{u} \).

Step 6 - Apply the Limits

Evaluate \( e^{u} \) at the new limits. This gives \( e^0 - e^1 = 1 - e\).

Key Concepts

Integration by Substitution

Integration by substitution is a technique used to simplify complex integrals. It is based on the chain rule of differentiation. This method often involves the following steps:

First, identify an inner function in the integrand and its derivative.
Then, substitute the inner function with a new variable, usually denoted as 'u'. This transforms the integral into a simpler form.

For example, in the integral \(\int_{5 \pi / 2}^{3 \pi} e^{\sin (q)} \cdot \cos (q) d q\), identify \(\sin(q)\) as the inner function and \(\cos(q)\) as its derivative:

• Set \(u = \sin(q)\)
• Thus, \(du = \cos(q) dq\)

This changes the integral to \(\int e^{u} du\)

The advantage is that \(\int e^u du\) is simpler to solve than the original integral. Finally, don’t forget to revert the substitution if necessary and change the limits of integration according to the new variable.

Integrals

Integrals are an essential part of calculus and are used to find areas, volumes, central points, and much more. There are two main types of integrals:

• **Definite integrals**: These compute the accumulation of quantities over a particular interval, providing a numerical value. They have upper and lower limits.
• **Indefinite integrals**: These represent a family of functions defined as antiderivatives. They do not have limits and result in a function plus a constant of integration (C).

In our exercise, we are dealing with a definite integral where we use the new limits:

The original integral was \(\int_{5 \pi / 2}^{3 \pi} e^{\sin (q)} \cdot \cos (q) d q\)
and it was transformed into \(\int_{1}^{0} e^{u} du\). Integrating this gives \(e^u\). Applying the limits of integration:
\(\left[ e^u \right]_1^0 = e^0 - e^1 = 1 - e\).

Calculus

Calculus is the branch of mathematics that studies how things change. It is divided mainly into two parts:

• **Differential Calculus**: Focuses on finding rates of change (derivatives).
• **Integral Calculus**: Concerned with accumulation of quantities and areas under or between curves (integrals).

The Fundamental Theorem of Calculus connects these two branches. It states that differentiation and integration are inverse processes. Specifically, to find the integral of a function, one must find its antiderivative and evaluate it over an interval.

In our problem, the theorem shows step-by-step how to evaluate the given definite integral. By first using substitution, we simplify the integration process and then use an antiderivative to find the exact result. This process highlights how integral calculus enables the calculation of areas and accumulations clearly and precisely.