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Magazine Chapter 5: Problem 4
Evaluate the integral. $$\int \frac{r^{2}}{r+4} d r$$
Short Answer
Expert verified
The integral evaluates to \( \frac{r^2}{2} - 4r + 16 \ln|r+4| + C \).
Step by step solution
01
Use Polynomial Long Division
To evaluate the integral \( \int \frac{r^2}{r+4} \, dr \), start with polynomial long division. Divide \( r^2 \) by \( r+4 \). The first term is \( r \), leading to \( r(r+4) = r^2 + 4r \). Subtract \( (r^2 + 4r) \) from \( r^2 \) to get \( -4r \).
02
Continue Polynomial Long Division
Divide \( -4r \) by \( r+4 \). The next term is \(-4\), yielding \(-4(r + 4) = -4r - 16 \). Subtracting \(-4r - 16\) from \(-4r\), we get the remainder \(16\). So, \( \frac{r^2}{r + 4} = r - 4 + \frac{16}{r+4} \).
03
Separate the Integral
Rewrite the integral using the result of the division: \( \int (r - 4 + \frac{16}{r+4}) \, dr \). This gives us \( \int r \, dr - \int 4 \, dr + \int \frac{16}{r+4} \, dr \).
04
Integrate Each Term
Integrate each term separately: \( \int r \, dr = \frac{r^2}{2} \), \( \int 4 \, dr = 4r \), and \( \int \frac{16}{r+4} \, dr = 16 \ln|r+4| \).
05
Combine the Results
Combine the integrals to get the final result: \( \frac{r^2}{2} - 4r + 16 \ln|r+4| + C \), 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.
Polynomial Long Division
Polynomial long division is a method used to divide a polynomial by another polynomial of lesser degree. This process is similar to the long division you use with numbers. It helps to simplify expressions into a form that鈥檚 easier to integrate.
Let鈥檚 see how it works through our exercise. We needed to divide the polynomial \( r^2 \) by \( r + 4 \). Here鈥檚 how we do it:
Let鈥檚 see how it works through our exercise. We needed to divide the polynomial \( r^2 \) by \( r + 4 \). Here鈥檚 how we do it:
- First, divide the leading term of the dividend (\( r^2 \)) by the leading term of the divisor (\( r \)). This gives you the first term of the quotient, \( r \).
- Multiply the divisor \( r + 4 \) by \( r \), giving \( r^2 + 4r \).
- Subtract \( r^2 + 4r \) from \( r^2 \), resulting in \( -4r \).
- Next, divide \( -4r \) by \( r \) to get the next term, \(-4\).
- Multiply \(-4\) by the divisor \( r + 4 \) and subtract it from \(-4r \). You will be left with 16 as a remainder.
Indefinite Integral
An indefinite integral represents a family of functions and is often used when dealing with integration problems. In simple terms, it asks for an antiderivative for a given function.
While working with indefinite integrals, breaking down the integrals into manageable parts like what we have in \( \int (r - 4 + \frac{16}{r+4}) \, dr \) simplifies the calculation. Here are the steps you can use:
While working with indefinite integrals, breaking down the integrals into manageable parts like what we have in \( \int (r - 4 + \frac{16}{r+4}) \, dr \) simplifies the calculation. Here are the steps you can use:
- Split the integral: \( \int r \, dr - \int 4 \, dr + \int \frac{16}{r+4} \, dr \).
- For \( \int r \, dr \), the antiderivative of \( r \) is \( \frac{r^2}{2} \).
- For \( \int 4 \, dr \), the antiderivative is \( 4r \).
- For \( \int \frac{16}{r+4} \, dr \), this can be solved using another core concept known as the natural logarithm, because it involves a reciprocal term \( \frac{1}{r+4} \).
Natural Logarithm
The natural logarithm, usually denoted as \( \ln(x) \), is a fundamental concept in calculus involving the integral of the reciprocal of \( x \), which is \( \int \frac{1}{x} \, dx = \ln|x| \) plus a constant.
When we came to \( \int \frac{16}{r+4} \, dr \), the natural logarithm concept is what we needed. Here鈥檚 how it worked:
When we came to \( \int \frac{16}{r+4} \, dr \), the natural logarithm concept is what we needed. Here鈥檚 how it worked:
- The \( 16 \) is a constant factor, so it can be pulled out of the integral: \( 16 \int \frac{1}{r+4} \, dr \).
- Using the natural logarithm rule, \( \int \frac{1}{u} \, du = \ln|u| + C \), where \( u = r+4 \), we find that \( \int \frac{1}{r+4} \, dr = \ln|r+4| \).
- Combine it back with the constant: \( 16 \ln|r+4| \) is the result.
