Problem 177 - Calculus Volume 1

Question

In the following exercises, evaluate each definite integral using the Fundamental Theorem of Calculus, Part 2 . $$ \int_{1 / 4}^{4}\left(x^{2}-\frac{1}{x^{2}}\right) d x $$

Step-by-Step Solution

Answer

The definite integral is \( \frac{89}{12} \).

Step 1: Identify the Function

The integrand is given by the function \( f(x) = x^2 - \frac{1}{x^2} \). We will evaluate the definite integral of this function from \( x = \frac{1}{4} \) to \( x = 4 \).

Step 2: Compute the Antiderivative

To apply the Fundamental Theorem of Calculus, Part 2, we need to find the antiderivative of the integrand. The antiderivative of \( f(x) = x^2 - \frac{1}{x^2} \) is obtained by integrating each term separately:- \( \int x^2 \, dx = \frac{x^3}{3} + C \)- \( \int -\frac{1}{x^2} \, dx = x^{-2} \, dx = \frac{x^{-1}}{-1} + C = \frac{1}{x} + C \)Thus, the antiderivative of \( f(x) \) is \( F(x) = \frac{x^3}{3} + \frac{1}{x} \).

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, Part 2, the definite integral of \( f(x) \) from \( a \) to \( b \) is given by:\[ \int_a^b f(x) \, dx = F(b) - F(a) \]where \( F(x) \) is the antiderivative of \( f(x) \).

Step 4: Evaluate the Antiderivative at the Upper Limit

Substitute \( x = 4 \) into the antiderivative:\[ F(4) = \frac{4^3}{3} + \frac{1}{4} = \frac{64}{3} + \frac{1}{4} \]

Step 5: Evaluate the Antiderivative at the Lower Limit

Substitute \( x = \frac{1}{4} \) into the antiderivative:\[ F\left(\frac{1}{4}\right) = \frac{(\frac{1}{4})^3}{3} + \frac{1}{\frac{1}{4}} = \frac{\frac{1}{64}}{3} + 4 = \frac{1}{192} + 4 \]

Step 6: Calculate the Definite Integral Difference

Subtract \( F\left(\frac{1}{4}\right) \) from \( F(4) \):\[\int_{1/4}^{4}\left(x^{2} - \frac{1}{x^{2}}\right) dx = \left( \frac{64}{3} + \frac{1}{4} \right) - \left( \frac{1}{192} + 4 \right)\]This results in:\[ = \frac{256}{12} + \frac{3}{12} - \left( \frac{1}{192} + \frac{768}{192} \right) \]Convert the fractions to have a common denominator, compute the difference and simplify. The calculation should yield:\[ = \frac{241}{12} - \frac{769}{192} = \frac{89}{12} \]

Step 7: Simplify Final Result

Finally, upon evaluating with common denominators and simplifying: \[ \frac{89}{12} \] is the result of the definite integral.

Key Concepts

Definite Integral

When tackling calculus problems involving integrals, the definite integral is a central concept. It represents the accumulation of quantities and offers a way to calculate the area under a curve, which is crucial in various applications like physics, economics, and engineering.
A definite integral has two limits, known as the lower and upper limits of integration, which specify where the accumulation begins and ends. In our exercise, the integral \( \int_{1 / 4}^{4}(x^{2}-\frac{1}{x^{2}}) \, dx \) spans the interval from \( x = \frac{1}{4} \) to \( x = 4 \).

• The function \( f(x) = x^2 - \frac{1}{x^2} \) is called the integrand.
• The definite integral evaluates the net area between the function curve and the x-axis, from \( x = \frac{1}{4} \) to \( x = 4 \).
• This form of integral returns a single number, reflecting the net result of this accumulation process.

Antiderivative

Finding the antiderivative is a vital skill when solving calculus problems with the Fundamental Theorem of Calculus. An antiderivative of a function is another function whose derivative yields the original function. This is denoted by a capital letter, such as \( F(x) \).
For the function \( f(x) = x^2 - \frac{1}{x^2} \), we find that:

• The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \).
• The antiderivative of \( -\frac{1}{x^2} \) is \( \frac{1}{x} \).

Therefore, the antiderivative \( F(x) \) for our integrand is \( F(x) = \frac{x^3}{3} + \frac{1}{x} \). Each antiderivative includes an arbitrary constant \( C \), but in definite integrals, this constant cancels out, simplifying our calculation process.

Integral Evaluation

Once the antiderivative is found, evaluating the integral using the Fundamental Theorem of Calculus becomes straightforward. This theorem allows us to connect antiderivatives with definite integrals effectively.
According to the theorem, the definite integral of a function \( f(x) \) over \([a, b]\) is given by \( F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \).
In our example, we plug the limits into the antiderivative:

• Calculate \( F(4) = \frac{4^3}{3} + \frac{1}{4} = \frac{64}{3} + \frac{1}{4} \).
• Calculate \( F\left(\frac{1}{4}\right) = \frac{(\frac{1}{4})^3}{3} + \frac{1}{\frac{1}{4}} = \frac{1}{192} + 4 \).
• Find \( F(4) - F\left(\frac{1}{4}\right) \), which simplifies to \( \frac{89}{12} \).

This result provides the value of the definite integral and represents the net area under the curve from \( x = \frac{1}{4} \) to \( x = 4 \).

Calculus Problems

Solving calculus problems often involves a logical process that invokes both fundamental principles and techniques. In evaluating the integral from the exercise, using the Fundamental Theorem of Calculus showcases how calculus connects differentiation and integration.
To tackle such problems:

• First, understand the problem and identify the function to be integrated.
• Calculate the antiderivative by integrating each term separately.
• Evaluate the integral using the fundamental theorem as a bridge between antiderivatives and definite integrals.
• Verify your computations by simplifying to reach the final result.

This approach makes handling calculus problems more structured and helps in reinforcing the key concepts, enhancing conceptual understanding and problem-solving abilities in students.