Problem 193 - Calculus Volume 1

Question

In the following exercises, use the evaluation theorem to express the integral as a function \(F(x)\) . $$ \int_{-x}^{x} \sin t d t $$

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Identify the bounds and the integrand

The exercise requires us to evaluate the definite integral \( \int_{-x}^{x} \sin t \, dt \). The integrand is \( \sin t \), and the limits of integration are from \(-x\) to \(x\).

Step 2: Find the antiderivative

The antiderivative of \( \sin t \) is \( -\cos t \). Therefore, for the function \( f(t) = \sin t \), the corresponding antiderivative is \( F(t) = -\cos t \).

Step 3: Apply the Evaluation Theorem

The Evaluation Theorem states that for \( F(t) \) being the antiderivative of \( f(t) \), the definite integral from \( a \) to \( b \) is given by \( F(b) - F(a) \). Here, it translates to \( F(x) - F(-x) \).

Step 4: Compute \( F(x) \) and \( F(-x) \)

Calculate \( F(x) = -\cos(x) \) and \( F(-x) = -\cos(-x) \). Since \( \cos(-x) = \cos(x) \), we have both \( F(x) = -\cos(x) \) and \( F(-x) = -\cos(x) \).

Step 5: Subtract \( F(-x) \) from \( F(x) \)

Taking \( F(x) - F(-x) = -\cos(x) - (-\cos(x)) = 0 \). Therefore, \( \int_{-x}^{x} \sin t \, dt = 0 \).

Key Concepts

Definite Integral

A definite integral represents the area under a curve determined by a function on a given interval. It is expressed in the form \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) is the integrand, and \( a \) and \( b \) are the limits of integration. This expression calculates the net area, taking into account regions lying above and below the x-axis.

• The definite integral is evaluated with the help of the Evaluation Theorem, which enables us to find the accumulated quantity between two points on a function chart.
• Definite integrals can be interpreted as a net signed sum of infinitesimal values, conveying not just the displaced area but values accounting for direction (positive above the axis, negative below).

This concept appears in cases where you need to determine cumulative quantities like total displacement or mass distribution over an interval.

Antiderivative

An antiderivative of a function is another function whose derivative results in the original function. Finding an antiderivative is often a key step in evaluating a definite integral. Consider, for example, if you have a function \( f(x) \), an antiderivative of \( f \) is something like \( F(x) \) such that \( F'(x) = f(x) \).

• When calculating a definite integral, the antiderivative is crucial. It is used to apply the Evaluation Theorem.
• In the exercise, the antiderivative of \( \sin t \) is \( -\cos t \).

This means if you differentiate \( -\cos t \), you'll get \( \sin t \), confirming that it's correct. This fact underlines the importance of knowing how to find antiderivatives for effective integration.

Trigonomtric Integrals

Trigonometric integrals involve functions like sine and cosine, which appear frequently in calculus problems. These integrals often require specific knowledge of trigonometric identities and derivative rules.

• For example, in our exercise, the integral of \( \sin t \) was considered.
• Understanding how to transform and apply trigonometric identities can be crucial for solving these integrals efficiently.

Knowing that the derivative of \( \cos(t) \) is \( -\sin(t) \), leads to finding the antiderivative of \( \sin(t) \), which is \( -\cos(t) \). Such manipulations and identities are part of mastering trigonometric integrals.

Symmetry in Integration

Symmetry can simplify the process of evaluating integrals. In the exercise, the integral evaluated from \(-x\) to \(x\) of the function \(\sin t\) results in zero because of its symmetry.

• Trigonometric functions like sine are often symmetric. The function \(\sin t\) is an odd function, meaning \(\sin(-t) = -\sin(t)\).
• As a result, if you integrate an odd function over a symmetric interval around zero, the contributions from equal distances in opposite directions cancel out.

This feature makes symmetry particularly helpful in simplifying integration calculations, efficiently leading to results without extensive computation.