Problem 37 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the definite integral. \(\int_{0}^{1} \cos (\pi t / 2) d t\)

Step-by-Step Solution

Answer

The definite integral evaluates to \( \frac{2}{\pi} \).

Step 1: Identify the Integral Objective

We are required to evaluate the definite integral \( \int_{0}^{1} \cos(\frac{\pi t}{2}) \, dt \). The objective is to find the area under the curve of \( \cos(\frac{\pi t}{2}) \) from \( t=0 \) to \( t=1 \).

Step 2: Find the Indefinite Integral

First, we find the antiderivative of \( \cos(\frac{\pi t}{2}) \). The antiderivative of \( \cos(kx) \) is \( \frac{1}{k} \sin(kx) \). Since \( k = \frac{\pi}{2} \), the antiderivative is:\[ F(t) = \frac{2}{\pi} \sin\left(\frac{\pi t}{2}\right) + C \] where \( C \) is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, we evaluate the definite integral by computing \( F(1) - F(0) \). Substitute \( t=1 \) and \( t=0 \) into the antiderivative:\[ F(1) = \frac{2}{\pi} \sin\left(\frac{\pi \cdot 1}{2}\right) = \frac{2}{\pi} \sin\left(\frac{\pi}{2}\right) = \frac{2}{\pi} \times 1 = \frac{2}{\pi} \]\[ F(0) = \frac{2}{\pi} \sin\left(\frac{\pi \cdot 0}{2}\right) = \frac{2}{\pi} \times 0 = 0 \]

Step 4: Calculate the Definite Integral

Subtract the two results obtained from the Fundamental Theorem of Calculus calculation:\[ F(1) - F(0) = \frac{2}{\pi} - 0 = \frac{2}{\pi} \]

Key Concepts

Antiderivative

An antiderivative is a fundamental concept in calculus, representing a function whose derivative is the original function. In simpler terms, if you differentiate the antiderivative, you'll get back the function you started with. Every function has infinitely many antiderivatives, differing by a constant. This is because the derivative of a constant is zero, thus any constant can be added to an antiderivative without changing its derivative result.

• The antiderivative is often noted by the presence of an arbitrary constant of integration, denoted as \(C\).
• For trigonometric functions like \(\cos(kx)\), it’s useful to know that the antiderivative becomes \(\frac{1}{k} \sin(kx)\), where \(k\) is a constant.

This idea is crucial for solving definite integrals, as it allows us to find the indefinite integral first, simplifying further calculations.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of derivatives and integrals, forming the backbone of integral calculus. It essentially states that if you have a continuous function \(f(x)\) over an interval \([a, b]\), then the integral of \(f\) over \([a, b]\) is given by finding the antiderivative \(F(x)\) and computing \(F(b) - F(a)\).

• This theorem is what allows us to compute definite integrals efficiently, reducing them to a subtraction problem.
• Notice how in the example, after finding the antiderivative of \(\cos(\frac{\pi t}{2})\), we simply evaluated it at the boundaries, \(t=1\) and \(t=0\), to find the exact area under the curve from \(t=0\) to \(t=1\).

Without the Fundamental Theorem of Calculus, calculating areas and accumulated quantities would be far more challenging.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are key components in calculus due to their periodic nature and smooth behavior, making them ideal candidates for integration and differentiation. They describe angular relationships and are frequently used in problems related to waves, oscillations, and other repetitive phenomena.

• The function \(\cos(\theta)\), where \(\theta\) is an angle, provides the x-coordinate of a point on the unit circle corresponding to that angle.
• Knowing the basic antiderivatives of trigonometric functions aids significantly in solving integrals. For \(\cos(x)\), the antiderivative is \(\sin(x)\), while for \(\sin(x)\), it is \(-\cos(x)\).

In our exercise, the trigonometric function \(\cos(\pi t / 2)\) required understanding and using its antiderivative to find the area under its curve over a specified interval.