Chapter 5: Problem 46
Evaluate the integrals in Exercises \(17-50\) $$ \int(x+5)(x-5)^{1 / 3} d x $$
Short Answer
Expert verified
\(\frac{3}{7}(x-5)^{7/3} + \frac{15}{4}(x-5)^{4/3} + C\)
Step by step solution
01
Expand the Product
First, distribute the term \((x-5)^{1/3}\) with \(x+5\). This gives:\[(x+5)(x-5)^{1/3} = x(x-5)^{1/3} + 5(x-5)^{1/3}\]
02
Integrate Separately
Now, integrate each term separately:For the first term, \(\int x(x-5)^{1/3} \, dx\), use the substitution method. Let \(u = x-5\), which gives \(du = dx\) and \(x = u + 5\). The integral becomes:\[\int (u + 5) u^{1/3} \, du = \int (u^{4/3} + 5u^{1/3}) \, du\]For the second term, \(\int 5(x-5)^{1/3} \, dx\), again substitute \(u = x-5\):\[\int 5u^{1/3} \, du\]
03
Evaluate the Integrals
Integrate each term:For \(\int u^{4/3} \, du\), the integral is:\[\frac{3}{7} u^{7/3} + C_1\]For \(\int 5u^{1/3} \, du\), the integral is:\[\frac{5 imes 3}{4} u^{4/3} + C_2 = \frac{15}{4} u^{4/3}\]Finally, for \(\int 5u^{1/3} \, du\):\[5 \times \frac{3}{4}u^{4/3} = \frac{15}{4}u^{4/3} + C_3\]
04
Combine and Substitute Back
Combine the results from Step 3, and substitute \(u = x-5\) back into the expression:\[\frac{3}{7}(x-5)^{7/3} + \frac{15}{4}(x-5)^{4/3} + C\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The Substitution Method is a powerful technique used in calculus to simplify complex integrals. It involves changing variables in the integral to make the integration process easier. Essentially, you replace a part of the integral with a new variable, often denoted as \( u \), and find a corresponding \( du \). This method transforms the original integral into a simpler form.
For example, consider the integral from the exercise: \( \int(x+5)(x-5)^{1/3} \, dx \). To simplify, you can let \( u = x-5 \). Thus, \( du = dx \) and \( x = u + 5 \). When you substitute these into the integral, it becomes:
For example, consider the integral from the exercise: \( \int(x+5)(x-5)^{1/3} \, dx \). To simplify, you can let \( u = x-5 \). Thus, \( du = dx \) and \( x = u + 5 \). When you substitute these into the integral, it becomes:
- \( \int (u + 5) u^{1/3} \, du = \int (u^{4/3} + 5u^{1/3}) \, du \)
Polynomial Expansion
Polynomial Expansion is a process used to multiply expressions that involve binomials or polynomials. By expanding the polynomial, we rewrite it in a form that is easier to manipulate for further operations like integration or differentiation.
In this exercise, the phrase \( (x+5)(x-5)^{1/3} \) represents the multiplication of a binomial and a term involving a fractional exponent. To expand, you multiply each part of \( x+5 \) by \( (x-5)^{1/3} \):
In this exercise, the phrase \( (x+5)(x-5)^{1/3} \) represents the multiplication of a binomial and a term involving a fractional exponent. To expand, you multiply each part of \( x+5 \) by \( (x-5)^{1/3} \):
- First, \( x(x-5)^{1/3} \) which remains as it is but prepares for integration.
- Second, \( 5(x-5)^{1/3} \), which also sets the stage for independent integration.
Definite and Indefinite Integrals
Integrals are a fundamental part of calculus, categorized mainly into definite and indefinite integrals. An indefinite integral, like the one in this exercise, provides a general form of the anti-derivative of a function, plus a constant \( C \).
The exercise involves finding an indefinite integral for expressions such as \( \int u^{4/3} \, du \) and \( \int 5u^{1/3} \, du \). The result is a general solution involving a constant \( C \) since the limits of integration are not specified.
For indefinite integrals, always remember:
The exercise involves finding an indefinite integral for expressions such as \( \int u^{4/3} \, du \) and \( \int 5u^{1/3} \, du \). The result is a general solution involving a constant \( C \) since the limits of integration are not specified.
For indefinite integrals, always remember:
- They represent a family of functions since the constant \( C \) can be any real number.
- The power rule for integration, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), is commonly used.