Problem 101 - Calculus for Biology and Medicine

Question

Evaluate the definite integrals. $$ \int_{1}^{8} x^{-2 / 3} d x $$

Step-by-Step Solution

Answer

The value of the definite integral is 3.

Step 1: Identify the Integral Form and Limits

The given integral is \( \int_{1}^{8} x^{-2/3} \, dx \) with the limits of integration from 1 to 8. This is a definite integral of the form \( \int_{a}^{b} f(x) \, dx \).

Step 2: Find the Antiderivative

To evaluate the integral, find the antiderivative of \( x^{-2/3} \). The antiderivative of \( x^n \) is \( \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \). Here \( n = -\frac{2}{3} \), so the antiderivative is \( \frac{x^{(-2/3)+1}}{(-2/3)+1} + C = \frac{x^{1/3}}{1/3} + C = 3x^{1/3} + C \).

Step 3: Apply the Limits of Integration

Use the Fundamental Theorem of Calculus to evaluate the definite integral: \( \int_{1}^{8} x^{-2/3} \, dx = \left[ 3x^{1/3} \right]_1^8 \). This means you will substitute the upper limit first and then the lower limit into the antiderivative.

Step 4: Substitute the Upper Limit

Substitute 8 into the antiderivative: \( 3 \times 8^{1/3} = 3 \times 2 = 6 \), because \( 8^{1/3} = 2 \).

Step 5: Substitute the Lower Limit

Substitute 1 into the antiderivative: \( 3 \times 1^{1/3} = 3 \times 1 = 3 \), because \( 1^{1/3} = 1 \).

Step 6: Calculate the Definite Integral

Subtract the value of the antiderivative at the lower limit from the value at the upper limit: \( 6 - 3 = 3 \).

Step 7: Conclusion

The value of the definite integral \( \int_{1}^{8} x^{-2/3} \, dx \) is 3.

Key Concepts

Antiderivative

Understanding the antiderivative is essential when solving definite integrals. Essentially, the antiderivative of a function gives us the process of reversing differentiation. If we're given a function:

• For example, if we have a function like \( x^n \), the antiderivative is found using the formula \( \frac{x^{n+1}}{n+1} + C \), as long as \( n eq -1 \).
• Here, \( C \) represents the constant of integration, which we include in indefinite integrals. However, in definite integrals, this constant cancels out during evaluation, so we do not include it in our final answer.

To illustrate, in our exercise we find the antiderivative of \( x^{-2/3} \) using this formula. This results in the antiderivative \( 3x^{1/3} \). Having a solid grasp of how to derive an antiderivative is fundamental in calculus, especially when you're working to solve integrals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is key to connecting differentiation with integration. It's broken into two main parts, but for definite integrals, we're focusing on the second part, which can be summarized as:

• If \( F(x) \) is an antiderivative of \( f(x) \) on an interval \([a, b]\), then:\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]
• This tells us that to solve a definite integral, we simply need to evaluate the antiderivative at the upper limit, then at the lower limit, and subtract the two results.

In the exercise, we use this process to solve \( \int_{1}^{8} x^{-2/3} \, dx \). By calculating \( F(8) - F(1) \) after finding the antiderivative, we arrive at the final integral value. This theorem thus makes it easy to compute the total accumulated value over an interval by simple subtraction, once you have the antiderivative.

Limits of Integration

When solving definite integrals, the limits of integration are the bounds \( a \) and \( b \). They represent the interval on the x-axis over which we are integrating the function. Therefore:

• In our problem, the limits of integration are 1 and 8. This implies we are interested in the area under the curve of the function \( x^{-2/3} \) from \( x = 1 \) to \( x = 8 \).
• The limits serve to provide the starting and ending points of the integration. It's these bounds that differentiate a definite from an indefinite integral.

To effectively compute the definite integral, we must evaluate the antiderivative at each of these limits. Specifically, plugging each of these values into the antiderivative and computing the difference yields the integral's value. Understanding these limits is crucial, as they define the range over which we apply the Fundamental Theorem of Calculus.