Problem 12 - Calculus Single Variable

Question

In Exercises \(11-20,\) calculate \(F(x)=\int_{a}^{x} f(t) d t\) $$ f(t)=\sin (t) \quad a=\pi / 4 $$

Step-by-Step Solution

Answer

\( F(x) = -\cos(x) + \frac{\sqrt{2}}{2} \)

Step 1: Understanding the Problem

We are asked to find the function \( F(x) \) which is the integral of \( f(t) = \sin(t) \) from \( a = \frac{\pi}{4} \) to \( x \). This integral will help us express \( F(x) \) in terms of \( x \).

Step 2: Integral of \( f(t) = \sin(t) \)

Find the indefinite integral of \( f(t) \). The integral of \( \sin(t) \) is \(-\cos(t) + C \), where \( C \) is the constant of integration.

Step 3: Setup the Definite Integral

Set up the definite integral expression: \[ F(x) = \int_{\pi/4}^{x} \sin(t)\, dt \] This represents the signed area under the curve \( \sin(t) \) from \( t = \frac{\pi}{4} \) to \( t = x \).

Step 4: Evaluate the Definite Integral

Using the Fundamental Theorem of Calculus, which states that \[ \int_{a}^{b} f(t) \,dt = F(b) - F(a) \] if \( F \) is an antiderivative of \( f \), we find: \[ F(x) = (-\cos(x) + C) - (-\cos\left(\frac{\pi}{4}\right) + C) \]which simplifies to: \[ F(x) = -\cos(x) + \cos\left(\frac{\pi}{4}\right) \]

Step 5: Simplify the Result

Simplify \( F(x) = -\cos(x) + \cos\left(\frac{\pi}{4}\right) \). Since \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \), we have: \[ F(x) = -\cos(x) + \frac{\sqrt{2}}{2} \] This is the final expression for \( F(x) \) in terms of \( x \).

Key Concepts

Integral Calculus

Integral calculus is the branch of calculus that focuses on the concept of integration. Integration is essentially the reverse operation to differentiation. If you think of differentiation as the process of finding the rate at which something changes, integration allows you to find the total amount accumulated from that rate of change.

There are two main types of integrals:

• Indefinite integrals, which do not provide a set value since they represent a family of functions and include the constant of integration, typically denoted as "C".
• Definite integrals, which provide a precise numerical value because they have upper and lower limits, indicating the area under a curve or between curves.

Both types of integrals play a significant role in various applications, from calculating areas and volumes to solving complex equations in physics and engineering.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a key principle that links differentiation and integration, two main concepts in calculus. It consists of two parts:

1. The first part states that if you have a continuous function \( f \) on an interval \([a, b]\), the function \( F \), defined by the integral \( F(x) = \int_{a}^{x} f(t) \, dt \), is an antiderivative of \( f \). This means that the derivative of \( F(x) \) with respect to \( x \) is \( f(x) \).

• \( F'(x) = f(x) \)

2. The second part says that if \( F \) is an antiderivative of \( f \) on \([a, b]\), then the definite integral from \( a \) to \( b \) of \( f \) is given by:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

This theorem is foundational because it allows us to evaluate definite integrals using antiderivatives, making the computation of areas and total values straightforward.

Definite Integral

A definite integral calculates the net area between the curve represented by a function \( f \) and the x-axis over a specific interval \[ a, b \]. This area might be positive or negative depending on whether the function is above or below the x-axis.

To visualize, the definite integral \( \int_{a}^{b} f(t) \, dt \) represents:

• The sum of the areas of small rectangles under the function \( f(t) \) from \( t = a \) to \( t = b \).
• If \( f(t) \) is positive over the interval, it sums the area above the x-axis, and if negative, below it.

This results in a specific value, which can be used in real-world applications to find things like distances traveled, areas between curves, and accumulated quantities.

Antiderivative

An antiderivative is a function whose derivative is the given function \( f(x) \). This means if you differentiate the antiderivative, you get back the original function. In a way, antiderivatives are the "reverse" of derivatives.

For example, given \( f(t) = \sin(t) \), an antiderivative is \( F(t) = -\cos(t) \) because the derivative of \(-\cos(t)\) is \( \sin(t) \).

When working with integrals, especially indefinite integrals, finding an antiderivative is crucial. We express this with the integral sign leading to the solution including a constant:

• \( \int f(x) \, dx = F(x) + C \)

Here, \( C \) is the constant of integration, accounting for all possible vertical shifts of the antiderivative graph.