Problem 188 - Calculus Volume 1

Question

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{1}^{2}\left(\frac{1}{t^{2}}-\frac{1}{t^{3}}\right) d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{8} \).

Step 1: Identify the Integrand

The given definite integral is \( \int_{1}^{2} \left( \frac{1}{t^{2}} - \frac{1}{t^{3}} \right) dt \). We need to evaluate this integral by finding its antiderivative.

Step 2: Find the Antiderivative

The antiderivative of \( \frac{1}{t^2} \) is \( -\frac{1}{t} \), and the antiderivative of \( \frac{1}{t^3} \) is \( -\frac{1}{2t^2} \). Thus, the antiderivative of the integrand \( \frac{1}{t^2} - \frac{1}{t^3} \) is: \[ F(t) = -\frac{1}{t} + \frac{1}{2t^2} \]

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, Part 2, the definite integral from 1 to 2 of a function is the difference between the values of its antiderivative at these points. Therefore, evaluate: \[ \int_{1}^{2} \left( \frac{1}{t^2} - \frac{1}{t^3} \right) dt = F(2) - F(1) \]

Step 4: Evaluate F(2) and F(1)

Calculate \( F(2) \): \( -\frac{1}{2} + \frac{1}{8} = -\frac{4}{8} + \frac{1}{8} = -\frac{3}{8} \). Calculate \( F(1) \): \( -1 + \frac{1}{2} = -\frac{2}{2} + \frac{1}{2} = -\frac{1}{2} \).

Step 5: Calculate the Definite Integral

Substitute \( F(2) \) and \( F(1) \) into the expression \( F(2) - F(1) \), which gives: \[ -\frac{3}{8} - \left(-\frac{1}{2}\right) = -\frac{3}{8} + \frac{4}{8} = \frac{1}{8} \]

Step 6: Conclusion: Evaluate the Integral

The value of the definite integral \( \int_{1}^{2} \left( \frac{1}{t^2} - \frac{1}{t^3} \right) dt \) is \( \frac{1}{8} \).

Key Concepts

Definite Integral

The idea of a definite integral is central in calculus. It represents the accumulation of quantities, such as area under a curve, between two specified limits, often written as \( a \) and \( b \). This is in contrast to an indefinite integral, which does not have specified limits. The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) is the function (or integrand) we are integrating.

• The definite integral calculates the net difference between the integrand's antiderivative evaluated at \( b \) and \( a \).
• In our example, the integral from 1 to 2 is represented as \( \int_{1}^{2} \left( \frac{1}{t^{2}} - \frac{1}{t^{3}} \right) dt \).
• This requires finding the antiderivative of the function within the given limits, revealing the accumulation of the function from \( t = 1 \) to \( t = 2 \).

Exploring definite integrals helps solve real-world problems involving continuous accumulation, such as computing displacement from velocity.

Antiderivative

The antiderivative, also known as the indefinite integral, is a function that "undoes" the operation of differentiation. It is the function \( F(x) \) whose derivative \( F'(x) \) equals the given function \( f(x) \).

• To find an antiderivative, we essentially reverse the derivative process for each part of the integrand.
• For example, given the terms \( \frac{1}{t^2} \) and \( \frac{1}{t^3} \), we find they correspond to \( -\frac{1}{t} \) and \( -\frac{1}{2t^2} \), respectively, as their antiderivatives.
• Thus, the antiderivative of \( \frac{1}{t^2} - \frac{1}{t^3} \) is \( F(t) = -\frac{1}{t} + \frac{1}{2t^2} \).

Antiderivatives are crucial for solving definite integrals, as they allow the application of the Fundamental Theorem of Calculus to evaluate the accumulation of the function over an interval.

Evaluation of Integrals

Evaluating an integral, especially a definite one, involves several clear steps. This process is essential for finding the exact value that a function accumulates over an interval.

• First, determine the antiderivative of the integrand. This represents the accumulation function for the original rate of change expressed by the integrand.
• Next, apply the Fundamental Theorem of Calculus. This theorem states that to compute \( \int_{a}^{b} f(x) \, dx \), you subtract the antiderivative of \( f(x) \) evaluated at \( a \) from its value at \( b \).
• For our exercise, this is shown by calculating \( F(2) \) and \( F(1) \), then finding the difference \( F(2) - F(1) \).

In our problem, we computed this to find that \( \int_{1}^{2} \left( \frac{1}{t^2} - \frac{1}{t^3} \right) dt = \frac{1}{8} \). Properly evaluating integrals involves attentively applying these concepts to ensure accurate results.