Problem 54 - Calculus Single Variable

Question

\(F(x)\) is a function of a variable \(x\) that appears in a limit (or in the limits) of integration of a given definite integral. Express \(F(x)\) explicitly by calculating the integral. $$ F(x)=\int_{\pi / 6}^{x} 2 \cos (t) d t $$

Step-by-Step Solution

Answer

\( F(x) = 2 \sin(x) - 1 \)

Step 1: Understanding the Integral

The problem asks us to express the function \( F(x) \) explicitly by calculating the definite integral \( \int_{\pi / 6}^{x} 2 \cos(t) \, dt \). This integral has a variable upper limit, \( x \), meaning the result will also be a function of \( x \).

Step 2: Antiderivative of the Integrand

The first step in integrating \( 2 \cos(t) \) is to find its antiderivative. The antiderivative of \( \cos(t) \) is \( \sin(t) \), so the antiderivative of \( 2 \cos(t) \) is \( 2 \sin(t) \).

Step 3: Evaluate the Definite Integral

With the antiderivative \( 2 \sin(t) \) found, we now evaluate the definite integral from \( t = \pi/6 \) to \( t = x \). This involves finding the difference \( 2 \sin(x) - 2 \sin(\pi/6) \).

Step 4: Calculation of the Sine Values

We know that \( \sin(\pi/6) = 1/2 \), therefore the expression \( 2 \sin(\pi/6) \) simplifies to \( 1 \).

Step 5: Final Expression for \( F(x) \)

Substituting back into the integral evaluation gives us \( F(x) = 2 \sin(x) - 1 \).

Key Concepts

Antiderivative

To solve an integral, one of the most important steps is finding the antiderivative of the function you are integrating. An antiderivative of a function is essentially a function whose derivative is the original function. For example, if you have a function like \(2 \cos(t)\), you're looking for a function that, when differentiated, gives you back \(2 \cos(t)\).For the cosine function, \( \cos(t)\), its antiderivative is \( \sin(t)\). This is because the derivative of \( \sin(t)\) is indeed \( \cos(t)\). Likewise, for our initial function \(2 \cos(t)\), we multiply the antiderivative of \( \cos(t) \) by 2, giving us an antiderivative of \(2 \sin(t)\).Finding the antiderivative is crucial in working with integrals, particularly when you are preparing to evaluate a definite integral, which requires you to know this antiderivative beforehand.

Variable Limit

When dealing with definite integrals, a variable limit is an important aspect to understand. A variable limit means that one or both limits of your integral can change depending on a variable, most commonly \(x\).In our example, the upper limit of integration is \(x\), which means the definite integral is not simply a number, but a function with its value dependent on \(x\). This makes the problem more dynamic and introduces the fundamental idea of how integrals can output functions, not just numbers.To evaluate such integrals, we follow the same steps as usual: find the antiderivative, substitute the variable upper limit and the constant lower limit, and take the difference between these two outcomes.

Function of a Variable

In integration, sometimes the result of an integral is a function of a variable, instead of a simple number. Particularly for our goal, \(F(x)\) is a function of the variable \(x\) that depends on the integral from \(\pi/6\) to \(x\) of \(2 \cos(t)\) with respect to \(t\).Evaluating this integral provides a new function, fulfilling mathematical expectations from the context of variable upper limits. It's important to conceptualize that the calculation of an integral provides a form of transformation from one function to another, in the domain of the given limits.This results in an explicit functional expression of \(F(x)\) which, in this case, turns out to be \(2\sin(x) - 1\), deeply tied to what value \(x\) holds.

Evaluate Definite Integrals

Evaluating definite integrals involves determining the actual value or expression of an integral over a specified interval. For definite integrals, this generally means taking the antiderivative of your function and then using it to compute values at the upper and lower limits of the interval.To illustrate, after determining that the antiderivative of \(2 \cos(t)\) is \(2 \sin(t)\), you'd plug in the upper and lower limits, substituting \(t\) with \(x\) and \(\pi/6\) respectively. The expression evaluates as follows:

• Calculate \(2 \sin(x)\) for the upper limit.
• Subtract \(2 \sin(\pi/6)\), which evaluates to 1.

Finally, the result \(F(x) = 2 \sin(x) - 1\) offers the evaluated expression of our definite integral, giving you the function related to the specified limits.