Problem 30 - Calculus Single Variable

Question

Evaluate the given definite integral by finding an antiderivative of the integrand and applying Theorem \(3 .\) $$ \int_{\pi / 3}^{3 \pi} \sin (x) d x $$

Step-by-Step Solution

Answer

The value of the integral is \(\frac{3}{2}\).

Step 1: Identify the problem

We need to evaluate the definite integral \(\int_{\pi/3}^{3\pi} \sin(x) \, dx\) by finding an antiderivative of the integrand \(\sin(x)\) and applying the Fundamental Theorem of Calculus (Theorem 3).

Step 2: Find the antiderivative

The antiderivative of \(\sin(x)\) is \(-\cos(x)\). Therefore, we have \(F(x) = -\cos(x)\) such that \(F'(x) = \sin(x)\).

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, we evaluate the integral by computing the values of the antiderivative at the upper and lower limits of the interval and finding the difference: \[\int_{\pi/3}^{3\pi} \sin(x) \, dx = F(3\pi) - F(\pi/3).\]

Step 4: Evaluate at the upper limit

Substitute \(3\pi\) into the antiderivative: \[-\cos(3\pi) = -(-1) = 1.\]

Step 5: Evaluate at the lower limit

Substitute \(\pi/3\) into the antiderivative: \[-\cos(\pi/3) = -\left(\frac{1}{2}\right) = -\frac{1}{2}.\]

Step 6: Calculate the difference

Compute the difference of the antiderivative values found in Steps 4 and 5: \[1 - (-\frac{1}{2}) = 1 + \frac{1}{2} = \frac{3}{2}.\]

Key Concepts

Antiderivative

Understanding antiderivatives is key to solving definite integrals. An antiderivative of a function is another function whose derivative gives the original function back.
In simpler terms, finding an antiderivative of a given function is like reverse-engineering the derivative process.

To illustrate, consider the derivative of \(-abla\). When we differentiate \(-abla\), we get \(\sin(x)\). So, \(-abla\) is an antiderivative of \(\sin(x)\).
Thus, the role of finding an antiderivative in calculus is crucial as it forms the basis for evaluating definite integrals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects antiderivatives and definite integrals. It's a cornerstone of calculus that offers a way to evaluate integrals efficiently.

In essence, this theorem states: if \(F(x)\) is an antiderivative of \(f(x)\), meaning \(F'(x) = f(x)\), then the integral from \(a\) to \(b\) of \(f(x)\) is simply \([F(b) - F(a)]\).

Here's how it applies:

• First, find an antiderivative of the function you are integrating.
• Second, evaluate this antiderivative at the upper limit of the integral.
• Third, evaluate it at the lower limit.
• Finally, subtract these two results to find the value of the definite integral.

This elegant process saves us from more complex computations, leveraging the relationship between differentiation and integration.

Trigonometric Integration

Trigonometric integration involves integrating functions that contain trigonometric expressions like \(\sin(x)\), \(-\) or others.
These are very common in calculus and can be approached methodically once you know some basic antiderivatives.

In the exercise, \(\sin(x)\) integrates to \(-\). Such relationships come from understanding the derivatives of trigonometric functions first, since differentiation and integration are inverse operations.

Practical steps for handling trigonometric integration include:

• Identifying the trigonometric function involved.
• Using known antiderivatives, like knowing that \(\sin(x)\) integrates to \(-\).
• Applying trigonometric identities when necessary to simplify the integrand.

This process makes evaluating trigonometric integrals less daunting and more intuitive.