Chapter 7: Problem 6
Find the area under the curve \(y=f(x)\) over the stated interval. $$f(x)=x^{-3 / 5} ;[1,4]$$
Short Answer
Expert verified
The area under the curve from 1 to 4 is approximately 1.85.
Step by step solution
01
Understand the Problem
We need to find the area under the curve of the function \( f(x)=x^{-3/5} \) over the interval \([1, 4]\). This requires evaluating a definite integral over this interval.
02
Set Up the Integral
To find the area under the curve, we set up the definite integral from 1 to 4 for the function. The integral we want to compute is:\[\int_{1}^{4} x^{-3/5} \, dx\]
03
Find the Antiderivative
Calculate the antiderivative of \( x^{-3/5} \). The general power rule for integration is applied:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]For \( n = -3/5 \), the antiderivative is:\[\frac{x^{-3/5 + 1}}{-3/5 + 1} = \frac{x^{2/5}}{2/5} = \frac{5}{2} x^{2/5}\]
04
Evaluate the Definite Integral
Apply the limits of integration to the antiderivative. We need to evaluate:\[\frac{5}{2} x^{2/5} \bigg|_{1}^{4} = \frac{5}{2} \left( 4^{2/5} - 1^{2/5} \right)\]First, calculate \( 4^{2/5} \) and \( 1^{2/5} \) to get:- \( 4^{2/5} = (2^2)^{2/5} = 2^{4/5} \)- \( 1^{2/5} = 1 \) since any number raised to any power of 1 is still 1.
05
Solve for the Definite Integral
Substitute the values back to get:\[\frac{5}{2} (2^{4/5} - 1)\]Now, compute the approximate value for \( 2^{4/5} \) using a calculator, which is approximately 1.7411, and solve:\[\frac{5}{2} (1.7411 - 1) = \frac{5}{2} (0.7411) = 1.85275\]
06
State the Final Answer
The area under the curve \( y = x^{-3/5} \) from \( x = 1 \) to \( x = 4 \) is approximately 1.85.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antiderivative
The antiderivative, also known as the indefinite integral, is a central concept in calculus that helps us reverse the process of differentiation. It involves finding a function whose derivative is the given function. To understand it better, think of it like this: if differentiation breaks down a function, the antiderivative builds it back up.
In our exercise, we sought the antiderivative of the function \( f(x) = x^{-3/5} \). The power rule helps us here (which we'll explore more in the next section). By applying this rule, you find the antiderivative of \( x^{-3/5} \) to be \( \frac{5}{2}x^{2/5} \). While the antiderivative provides a family of functions (differing by a constant \( C \)), in definite integral applications, we focus on evaluating at specific bounds, so \( C \) cancels out.
For anyone struggling with this concept, just remember: if you integrate a derivative, you return to the function you began with, only without specific constants in the case of definite integrals. This ability to "reverse" differentiation makes antiderivatives incredibly useful for solving real-world problems like finding areas under curves.
In our exercise, we sought the antiderivative of the function \( f(x) = x^{-3/5} \). The power rule helps us here (which we'll explore more in the next section). By applying this rule, you find the antiderivative of \( x^{-3/5} \) to be \( \frac{5}{2}x^{2/5} \). While the antiderivative provides a family of functions (differing by a constant \( C \)), in definite integral applications, we focus on evaluating at specific bounds, so \( C \) cancels out.
For anyone struggling with this concept, just remember: if you integrate a derivative, you return to the function you began with, only without specific constants in the case of definite integrals. This ability to "reverse" differentiation makes antiderivatives incredibly useful for solving real-world problems like finding areas under curves.
Power Rule for Integration
The power rule for integration is a powerful tool that simplifies finding antiderivatives of functions in polynomial form. It states that when integrating a function \( x^n \), the result is \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), given \( n eq -1 \). This rule directly informs how we approach the integration of polynomials—making calculations quicker and easier.
For the function \( x^{-3/5} \), the power rule allows us to find the antiderivative. We identify \( n = -3/5 \) and proceed according to the rule, incrementing the exponent by 1 and dividing by the new exponent. The calculated antiderivative was \( \frac{5}{2}x^{2/5} \), later used to solve the definite integral.
Keep in mind:
For the function \( x^{-3/5} \), the power rule allows us to find the antiderivative. We identify \( n = -3/5 \) and proceed according to the rule, incrementing the exponent by 1 and dividing by the new exponent. The calculated antiderivative was \( \frac{5}{2}x^{2/5} \), later used to solve the definite integral.
Keep in mind:
- The power rule applies to polynomials, so ensure you rewrite functions as needed, such as turning roots into fractional exponents.
- Always adjust for indefinite integrals with \( + C \) to account for constant terms.
- When using this rule in definite integrals, focus on evaluating the antiderivative at the upper and lower bounds.
Area under the curve
Finding the area under a curve is one of the most common applications of definite integrals. This process allows us to determine the net area between the curve of a function and the x-axis over a specific interval. The term "net" is important because areas below the x-axis subtract from those above.
For the problem at hand, we needed to calculate the area under \( y = x^{-3/5} \) from \( x = 1 \) to \( x = 4 \). Using the definite integral \( \int_{1}^{4} x^{-3/5} \, dx \), we integrate and use the bounds to find the actual numeric area. The result is approximately 1.85, a measure of the space between the curve and the x-axis over our interval.
Key points to remember:
For the problem at hand, we needed to calculate the area under \( y = x^{-3/5} \) from \( x = 1 \) to \( x = 4 \). Using the definite integral \( \int_{1}^{4} x^{-3/5} \, dx \), we integrate and use the bounds to find the actual numeric area. The result is approximately 1.85, a measure of the space between the curve and the x-axis over our interval.
Key points to remember:
- A positive result implies the curve remains above the x-axis within the interval.
- For intervals across the x-axis, areas below it contribute negative values, affecting the net result.
- Deploying numerical methods (or calculators) aids significantly when dealing with complex functions.