Problem 53 - Calculus: Early Transcendentals

Question

Evaluate the definite integral. \( \displaystyle \int^1_0 \cos(\pi t/2) \, dt \)

Step-by-Step Solution

Answer

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

Step 1: Set Up the Integral

We need to evaluate the definite integral \( \int^1_0 \cos(\pi t/2) \, dt \). This means we'll find the integral of \( \cos(\pi t/2) \) from \( t = 0 \) to \( t = 1 \).

Step 2: Find the Antiderivative

The function to integrate is \( f(t) = \cos(\pi t/2) \). Find an antiderivative for this function. Since the derivative of \( \sin(\pi t/2) \) is \( \frac{\pi}{2} \cdot \cos(\pi t/2) \), we need \( \sin(\pi t/2) \) divided by \( \pi/2 \). Thus, the antiderivative is \( \frac{2}{\pi} \sin(\pi t/2) \).

Step 3: Apply the Fundamental Theorem of Calculus

By the Fundamental Theorem of Calculus, evaluate the antiderivative \( \frac{2}{\pi} \sin(\pi t/2) \) at the bounds of integration: \[ \left[ \frac{2}{\pi} \sin(\pi t/2) \right]^1_0 = \frac{2}{\pi} \sin(\pi/2) - \frac{2}{\pi} \sin(0). \]

Step 4: Simplify and Calculate

Calculate the values: \( \sin(\pi/2) = 1 \) and \( \sin(0) = 0 \). So the expression simplifies to \( \frac{2}{\pi} \cdot 1 - \frac{2}{\pi} \cdot 0 = \frac{2}{\pi}. \)

Key Concepts

Understanding the Antiderivative

In calculus, an antiderivative is a function that takes us back to the original function when differentiated. Essentially, it's the reverse of differentiation.
When we say we are finding the antiderivative of a function, we are looking for a new function whose derivative gives us our original function back. This process is central to solving integrals. For example, in our exercise, we needed the antiderivative of \( \cos(\pi t/2) \). We know that the derivative of \( \sin(u) \) is \( \cos(u) \), so logically, \( \sin(\pi t/2) \) must be the antiderivative of \( \cos(\pi t/2) \), especially after adjusting for the constant \( \frac{\pi}{2} \).
Subsequently, the antiderivative becomes \( \frac{2}{\pi} \sin(\pi t/2) \), ensuring that it compensates for the coefficient within the cosine's argument.

Trigonometric Integration

Trigonometric integration involves integrating trigonometric functions like sine, cosine, tangent, and others.
These functions often appear in calculus problems due to their oscillatory nature and periodic properties.When integrating a function like \( \cos(\pi t/2) \), recognizing the basic integral form is essential.
In our problem, we note that the integral of a simple cosine function \( \cos(u) \) is just \( \sin(u) \). However, the argument in our integral's cosine, \( \pi t/2 \), is a little more complex, requiring a substitution or recognition of transformation. We needed to adjust the antiderivative by dividing by the derivative of the inner function, \( \frac{\pi}{2} \), which we do by multiplying by its reciprocal \( \frac{2}{\pi} \).
Such adjustments are common when dealing with trigonometric integrals that involve more complex arguments.

Applying the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of differentiation with integration. It tells us that if we have an antiderivative of a function, we can evaluate a definite integral simply by focusing on this antiderivative at specified bounds: the upper and lower limits.
In simpler terms, it helps compute the area under a curve over an interval.For our exercise, once the antiderivative \( \frac{2}{\pi} \sin(\pi t/2) \) was determined, the theorem allowed us to take its values at \( t = 1 \) and \( t = 0 \).
This meant substituting \( t = 1 \) into \( \frac{2}{\pi} \sin(\pi t/2) \) and then subtracting the value of the antiderivative at \( t = 0 \).Applying this, we calculated the integral to be \( \frac{2}{\pi} \), as \( \sin(\pi/2) = 1 \) and \( \sin(0) = 0 \).
This demonstrates the effectiveness of the Fundamental Theorem in providing a straightforward way to determine integrals.