Chapter 5: Problem 54
Evaluate the integrals using appropriate substitutions. $$\int x \sqrt{4-x} d x$$
Short Answer
Expert verified
The integral is \(-\frac{8}{3}(4-x)^{3/2} + \frac{2}{5}(4-x)^{5/2} + C\).
Step by step solution
01
Choose a Substitution
To simplify the integral \( \int x \sqrt{4-x} \, dx \), we choose a substitution that can simplify the expression inside the integral. Let's use the substitution \( u = 4 - x \). Consequently, \( du = -dx \). Thus, \( dx = -du \). Also, rewrite \( x \) in terms of \( u \): since \( u = 4 - x \), it follows that \( x = 4 - u \).
02
Substitute in the Integral
Substitute the expressions for \( x \) and \( dx \) into the integral. The integral becomes: \[ \int x \sqrt{4-x} \, dx = \int (4 - u) \sqrt{u} \, (-du) = \int -(4-u) u^{1/2} \, du. \]
03
Simplify and Rearrange
Simplify the integral expression. The minus sign can be brought outside the integral, leading to: \[ -\int (4 - u) u^{1/2} \, du = -\int (4u^{1/2} - u^{3/2}) \, du. \] Split the integral into two separate integrals: \[ = -\int 4u^{1/2} \, du + \int u^{3/2} \, du. \]
04
Integrate Term by Term
Now, integrate each term separately:1. Compute \( \int 4u^{1/2} \, du \): \[ \int 4u^{1/2} \, du = 4 \cdot \frac{u^{3/2}}{3/2} = \frac{8}{3} u^{3/2}. \] 2. Compute \( \int u^{3/2} \, du \): \[ \int u^{3/2} \, du = \frac{u^{5/2}}{5/2} = \frac{2}{5} u^{5/2}. \]
05
Combine Results
Combine the results of the integrations in Step 4:\[-\left( \frac{8}{3} u^{3/2} \right) + \left( \frac{2}{5} u^{5/2} \right)\]Simplify this to:\[-\frac{8}{3} u^{3/2} + \frac{2}{5} u^{5/2}.\]
06
Resubstitute Original Variable
Replace \( u \) with the original variable using the substitution \( u = 4 - x \). The final result in terms of \( x \) is:\[-\frac{8}{3} (4-x)^{3/2} + \frac{2}{5} (4-x)^{5/2} + C,\]where \( C \) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals are a fundamental concept in calculus, providing a way to compute accumulated quantities, such as area, when the boundaries of integration are fixed. When dealing with definite integrals, you're essentially finding the accumulation of a function's values over a certain interval.
In definite integrals, you have:
Ultimately, evaluating definite integrals gives you a numerical value representing the area under the curve of a given function between two points.
In definite integrals, you have:
- Upper and lower limits, which define the interval you're interested in.
- The notation typically looks like this: \(\int_{a}^{b} f(x) \, dx\), where \(a\) and \(b\) are the bounds of integration.
Ultimately, evaluating definite integrals gives you a numerical value representing the area under the curve of a given function between two points.
Antiderivatives
Antiderivatives, often called indefinite integrals, are the reverse process of differentiation. When you find an antiderivative, you're essentially looking for a function whose derivative is the original function. In mathematical notation, if you're given a function \(f(x)\), an antiderivative is denoted \(F(x)\), where \(F'(x) = f(x)\).
In practical terms, finding an antiderivative involves:
They play a crucial role in solving many integral problems, particularly when moving from abstract concepts to concrete numerical results in applications.
In practical terms, finding an antiderivative involves:
- Identifying a function \(F(x)\) such that its derivative returns the function \(f(x)\) within the integral.
- Introducing a constant of integration, \(C\), because indefinite integration isn't bound by specific limits.
They play a crucial role in solving many integral problems, particularly when moving from abstract concepts to concrete numerical results in applications.
U-Substitution
U-substitution is an integration technique used to simplify integrals by choosing a substitution that makes the integral easier to solve. It resembles the chain rule in reverse and is particularly useful when dealing with composite functions.
The method involves:
This substitution method often makes a complicated integral much simpler by reducing the complexity of the function or expression you need to integrate.
The method involves:
- Choosing a substitution \(u = g(x)\) that simplifies part of the integral.
- Rewriting the differential \(dx\) in terms of \(du\), usually by finding \(du = g'(x)dx\).
- Transforming the integral into terms of \(u\) and \(du\), making it easier to integrate.
This substitution method often makes a complicated integral much simpler by reducing the complexity of the function or expression you need to integrate.