Problem 6 - Calculus

Question

Use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{1}^{3} \frac{2}{t^{3}} d t $$

Step-by-Step Solution

Answer

The value of the definite integral is \( \frac{8}{9} \).

Step 1: Identify the integrand

The integrand given is \( \frac{2}{t^3} \). This can be rewritten as \( 2t^{-3} \).

Step 2: Find the antiderivative

To find the antiderivative of \( 2t^{-3} \), we use the power rule for integration: for \( t^n \), the antiderivative is \( \frac{t^{n+1}}{n+1} \). So the antiderivative of \( 2t^{-3} \) is \( 2 \cdot \frac{t^{-2}}{-2} = -t^{-2} \).

Step 3: Evaluate the antiderivative at the bounds

Apply the Second Fundamental Theorem of Calculus. Evaluate the antiderivative, \(-t^{-2}\), at the upper limit, \(t=3\), and the lower limit, \(t=1\).

Step 4: Calculate \(-t^{-2}\) at \(t=3\)

Substitute \(t=3\) into \(-t^{-2}\): \(-3^{-2} = -\frac{1}{9}\).

Step 5: Calculate \(-t^{-2}\) at \(t=1\)

Substitute \(t=1\) into \(-t^{-2}\): \(-1^{-2} = -1\).

Step 6: Find the difference

Subtract the value of the antiderivative at the lower limit from the value at the upper limit: \(-\frac{1}{9} - (-1) = -\frac{1}{9} + 1 = \frac{8}{9}\).

Key Concepts

Definite Integrals

Definite integrals are a key concept in calculus, playing a vital role in finding the area under a curve. When working with definite integrals, we're looking at integrals with specific limits or boundaries that define the range over which we integrate. In this case, the bounds are from 1 to 3.

The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration, respectively. The function \( f(x) \) is the integrand, essentially the curve you're measuring the area under. When you perform a definite integral, it results in a numerical value, unlike indefinite integrals which include a constant of integration.

• Definite integrals have both a direction and an interval.
• If you reverse the limits, you change the sign of the integral.
• If both limits are the same, the integral evaluates to zero, as there's no area to calculate.

Understanding these properties helps you work with definite integrals effectively and solve complex calculus problems with ease.

Integration Techniques

Various techniques are used to solve integrals, with each technique being handy in different situations. In the given exercise, we use the standard method known as the power rule for integration, which is especially useful when dealing with powers of variables.

For functions in the form of \( t^n \), the antiderivative can be found using \( \frac{t^{n+1}}{n+1} \). This simple formula helps find the antiderivative of many polynomial expressions quickly and efficiently. Here's why increasing the exponent by one during integration works:

• Integration is essentially the reverse process of differentiation.
• The power rule for integration undoes the power rule for differentiation.
• It allows us to simplify polynomial expressions and find their antiderivatives easily.

Other techniques like substitution and integration by parts are also available for more complex integrands. Becoming familiar with these techniques gives flexibility in solving a wide range of integrals.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are fundamental to understanding calculus, as they lay the groundwork for finding definite integrals through the Fundamental Theorem of Calculus. An antiderivative of a function \( f(t) \) is another function \( F(t) \) so that the derivative \( F'(t) \) is equal to \( f(t) \).

In our exercise, we needed the antiderivative of \( 2t^{-3} \). By applying the power rule, we identified the antiderivative as \( -t^{-2} \).

• Finding an antiderivative is a critical step before evaluating definite integrals.
• Antiderivatives include a generic constant \( C \), but definite integrals eliminate this constant when evaluated between limits.
• This process converts the complex behavior of differentiation into a tool for calculating exact areas under curves.

Mastering antiderivatives is the first step in effectively solving calculus problems that involve integration and understanding the behavior of functions over specific intervals.