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Magazine Chapter 2: Problem 32
Express \(x^{2}(a x+b)^{-1}\) as the sum of powers of \(x\) and another integrable term, and hence evaluate $$ \int_{0}^{b / a} \frac{x^{2}}{a x+b} d x $$
Short Answer
Expert verified
\int_{0}^{b/a} \frac{x^2}{a x + b} dx = -\frac{b^2}{2a^3} \.
Step by step solution
01
- Rewrite the integrand
Rewrite the integrand \( \frac{x^2}{a x + b} \) as \( x^2 \times (a x + b)^{-1} \).
02
- Use polynomial long division
Perform polynomial long division of \( x^2 \) by \( a x + b \) to express \( \frac{x^2}{a x + b} \) in a different form. This gives \( \frac{x^2}{a x + b} = \frac{1}{a}(x - \frac{b}{a}) + \frac{\frac{b^2}{a^2}}{a x + b} \).
03
- Simplify the expression
Rewrite the result from the polynomial division: \( \frac{1}{a}(x - \frac{b}{a}) + \frac{\frac{b^2}{a^2}}{a x + b} \).
04
- Set up the integral with the new expression
Rewrite the integral using the simplified expression. The integral becomes: \[ \begin{aligned} \int_{0}^{b/a} \frac{x^2}{a x + b} dx & = \frac{1}{a} \int_{0}^{b/a} (x - \frac{b}{a}) dx + \frac{b^2}{a^3} \int_{0}^{b/a} \frac{1}{a x + b} dx \end{aligned} \].
05
- Evaluate the first integral
Evaluate the first integral: \[ \frac{1}{a} \int_{0}^{b/a} (x - \frac{b}{a}) dx = \frac{1}{a} \[ \frac{x^2}{2} - \frac{b}{a}x \]_0^{b/a} \]. Plugging in the limits, it becomes: \frac{1}{a} \left( \frac{(b/a)^2}{2} - \frac{b}{a} \frac{b}{a} \right) = \frac{1}{a} \left( \frac{b^2}{2a^2} - \frac{b^2}{a^2} \right) = \frac{1}{a} \left( -\frac{b^2}{2a^2} \right) = \frac{-b^2}{2a^3} \.
06
- Evaluate the second integral
The second integral is: \frac{b^2}{a^3} \int_{0}^{b/a} \frac{1}{a x + b} dx = \frac{b^2}{a^3} \left[ \frac{1}{a} \ln \left( a x + b \right) \right]_0^{b/a} = \frac{b^2}{a^4} \[ \ln \left( b \right) - \ln \left( b \right) \]. Since both terms in the bracket cancel out, this part evaluates to 0.
07
- Sum the results
Combine the evaluated integrals from steps 5 and 6: \[ \frac{-b^2}{2a^3} + 0 = -\frac{b^2}{2a^3} \].
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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 similar to arithmetic long division. We use it to divide one polynomial by another. In our problem, we divide the polynomial \( x^2 \) by \( ax + b \). This simplifies our integral. Here's how it works:
First, we write the division as \( \frac{x^2}{ax+b} \).
We then perform the division step by step to get a polynomial quotient and a remainder. In this case, the division results in: \( x - \frac{b}{a} + \frac{\frac{b^2}{a^2}}{ax+b} \). This new form is easier to integrate!
First, we write the division as \( \frac{x^2}{ax+b} \).
We then perform the division step by step to get a polynomial quotient and a remainder. In this case, the division results in: \( x - \frac{b}{a} + \frac{\frac{b^2}{a^2}}{ax+b} \). This new form is easier to integrate!
Rational functions
A rational function is a ratio of two polynomials, like \( \frac{x^2}{ax+b} \). These functions often appear in calculus problems involving integrals. Using polynomial long division can simplify these ratios. This makes integration more manageable. Here, after performing polynomial long division, our rational function became: \(\frac{x^2}{ax+b} = \frac{1}{a}(x - \frac{b}{a}) + \frac{\frac{b^2}{a^2}}{ax+b}\). This form splits the function into simpler parts that we can integrate individually.
Definite integrals
Definite integrals calculate the area under a curve between two points. In our problem, we need to evaluate:
\( \int_{0}^{b/a} \frac{x^2}{ax+b} dx \).
Using the simplified form from polynomial long division, we rewrite the integral:
\( \int_{0}^{b/a} \frac{x^2}{ax+b} dx = \frac{1}{a} \int_{0}^{b/a} (x - \frac{b}{a}) dx + \frac{b^2}{a^3} \int_{0}^{b/a} \frac{1}{ax+b} dx \).
This breaks the integral into two easier parts: one polynomial and one rational function.
\( \int_{0}^{b/a} \frac{x^2}{ax+b} dx \).
Using the simplified form from polynomial long division, we rewrite the integral:
\( \int_{0}^{b/a} \frac{x^2}{ax+b} dx = \frac{1}{a} \int_{0}^{b/a} (x - \frac{b}{a}) dx + \frac{b^2}{a^3} \int_{0}^{b/a} \frac{1}{ax+b} dx \).
This breaks the integral into two easier parts: one polynomial and one rational function.
Logarithmic integration
Logarithmic integration involves integrating functions that result in logarithms. For our second integral part, we have:
\( \frac{b^2}{a^3} \int_{0}^{b/a} \frac{1}{ax+b} dx \).
The integral of \( \frac{1}{ax+b} \) is a natural logarithm:\( \int \frac{1}{ax+b} dx = \frac{1}{a} \ln(ax+b) \).
Using this, we evaluate the definite integral:
\( \frac{b^2}{a^3} \left[ \frac{1}{a} \ln(ax+b) \right]_0^{b/a} \), which eventually cancels out. This makes the integration result equal to 0 for this part.
By combining both results, we get the solution to the original problem.
\( \frac{b^2}{a^3} \int_{0}^{b/a} \frac{1}{ax+b} dx \).
The integral of \( \frac{1}{ax+b} \) is a natural logarithm:\( \int \frac{1}{ax+b} dx = \frac{1}{a} \ln(ax+b) \).
Using this, we evaluate the definite integral:
\( \frac{b^2}{a^3} \left[ \frac{1}{a} \ln(ax+b) \right]_0^{b/a} \), which eventually cancels out. This makes the integration result equal to 0 for this part.
By combining both results, we get the solution to the original problem.
