Problem 17 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in Problems \(1-20\) exactly. $$\int_{-1}^{1} \cos t d t$$

Step-by-Step Solution

Answer

The value of the integral is \(2\sin(1)\).

Step 1: Identify the Integrand

The problem requires evaluating the definite integral of the function \( f(t) = \cos t \) over the interval \([-1, 1]\).

Step 2: Antiderivative of the Integrand

The Fundamental Theorem of Calculus involves finding an antiderivative of the integrand. The antiderivative of \( \cos t \) is \( \sin t \) because the derivative of \( \sin t \) is \( \cos t \).

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(t) \) is an antiderivative of \( f(t) \), then the definite integral of \( f(t) \) from \( a \) to \( b \) is given by \( F(b) - F(a) \). In this case, \( F(t) = \sin t \), so the integral is \( \sin(1) - \sin(-1) \).

Step 4: Evaluate the Antiderivative at the Bounds

Evaluate \( F(t) = \sin t \) at the upper bound, 1: \( \sin(1) \). Then, evaluate \( F(t) \) at the lower bound, -1: \( \sin(-1) \).

Step 5: Substitute and Simplify the Equation

Substitute the evaluated values back into the expression: \( \sin(1) - \sin(-1) \). Knowing that \( \sin(-x) = -\sin(x) \), we rewrite \( \sin(-1) \) as \(-\sin(1)\). This simplifies to \( \sin(1) + \sin(1) = 2\sin(1) \).

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, bridging the gap between differentiation and integration through the Fundamental Theorem of Calculus. When evaluating a definite integral, such as \( \int_{-1}^{1} \cos t\, dt \), the goal is to find the signed area under the curve of the function \( f(t) = \cos t \) between the limits of integration, which are -1 and 1 in this case.
The process of evaluating definite integrals involves several steps:

• First, you need to identify the integrand, which is the function being integrated. Here, it is \( \cos t \).
• Next, you'll need an antiderivative of this function to apply the Fundamental Theorem of Calculus.
• The result represents the net area between the function and the x-axis, with areas above the x-axis being positive and areas below being negative.

Definite integrals are essential not only for finding areas but also for solving problems in physics and engineering, where they help us derive quantities like work and fluid force.

Antiderivative Calculus

Antiderivative calculus is integral to solving problems involving integration. An antiderivative of a function \( f(t) \) is another function \( F(t) \) whose derivative is \( f(t) \). In simpler terms, if you differentiate \( F(t) \), you end up with the original function \( f(t) \). Finding an antiderivative is crucial because it allows you to evaluate definite integrals using the Fundamental Theorem of Calculus.

For example, in the problem \( \int_{-1}^{1} \cos t\, dt \), we looked for an antiderivative of \( \cos t \). We know from our calculus knowledge that the derivative of \( \sin t \) is \( \cos t \), making \( \sin t \) an antiderivative of \( \cos t \).

Once an antiderivative is identified, integrating over a specific interval \([a, b]\) becomes a matter of substituting the upper and lower bounds into \( F(t) \) and calculating the difference \( F(b) - F(a) \). Understanding antiderivatives is key to mastering integration, as it allows one to reverse differentiation and find total accumulated change.

Trigonometric Integration

Trigonometric integration involves integrating functions that include trigonometric functions like sine, cosine, and tangent. These functions frequently appear in calculus problems due to their periodic nature and applications in modeling waves or oscillatory motions in physics.

In our original exercise, we dealt with integrating \( \cos t \), a trigonometric function. A helpful strategy with trigonometric integration is knowing or recognizing the derivatives of basic trigonometric functions.

For example:

• The derivative of \( \sin t \) is \( \cos t \), so \( \sin t \) serves as the antiderivative of \( \cos t \).
• Similarly, the derivative of \( \cos t \) is \(-\sin t\).

This knowledge is crucial because in trigonometric integration, recognizing these relationships allows you to quickly find antiderivatives and evaluate integrals.
With the exercise, evaluating the definite integral required recognizing the antiderivative \( \sin t\) of \( \cos t \), followed by applying the Fundamental Theorem of Calculus to simplify and solve the integral. Trigonometric integration is essential in solving a variety of real-world problems, from calculating angles to determining oscillations.