Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 7: Problem 73
Evaluate the integrals in Exercises \(67-80\) $$ \int_{1}^{2} \frac{8 d x}{x^{2}-2 x+2} $$
Short Answer
Expert verified
The integral evaluates to \(2\pi\).
Step by step solution
01
Simplify the Denominator
Start by making the substitution \( u = x^2 - 2x + 2 \). First, check if the denominator can be simplified by completing the square. Notice that \( x^2 - 2x + 2 \) can be rewritten by completing the square as \((x-1)^2 + 1\).
02
Perform Substitution
Let \( u = (x-1)^2 + 1 \). Therefore, \( du = 2(x-1) \, dx \), which implies \( dx = \frac{du}{2(x-1)} \). But first, we need to modify our integral expression based on the existing terms; however, substitution directly with \((x-1)^2 + 1\) simplifies finding the integral.
03
Substitute Variables and Limits
Substituting \( x = 1 + t \) simplifies the denominator: \((x-1)^2 + 1 = t^2 + 1\). The integral in terms of \( t \) becomes: \( \int_{0}^{1} \frac{8 \, dt}{t^2 + 1} \). The limits change according to our new variable \( t \) such that when \( x = 1, t = 0 \), and when \( x = 2, t = 1 \).
04
Evaluate the New Integral
The integral \( \int \frac{8}{t^2 + 1} dt \) becomes \( 8 \int \frac{dt}{t^2 + 1} \). The antiderivative of \( \frac{1}{t^2 + 1} \) is \( \arctan(t) \), so the integral becomes \( 8 \arctan(t) \).
05
Apply Limits and Finalize Solution
Evaluate the result \( 8 \arctan(t) \) from \( t = 0 \) to \( t = 1 \). This yields: \( 8 (\arctan(1) - \arctan(0)) \) which is \( 8 \left( \frac{\pi}{4} - 0 \right) = 2\pi \).
06
Conclusion
Thus, the value of the integral \( \int_{1}^{2} \frac{8 \, dx}{x^2 - 2x + 2} \) is \( 2\pi \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
Integration by substitution is a popular technique in calculus used to evaluate integrals. It's especially helpful when dealing with complex expressions.
The idea is to substitute part of the integral with a new variable to make the integration easier.
For example, consider the original integral \[ \int \frac{8 \, dx}{x^2 - 2x + 2} \].
The idea is to substitute part of the integral with a new variable to make the integration easier.
For example, consider the original integral \[ \int \frac{8 \, dx}{x^2 - 2x + 2} \].
- Identify a part of the function to replace. Here, we look at the denominator \( x^2 - 2x + 2 \).
- By completing the square, it becomes \( (x-1)^2 + 1 \).
- Let \( u = (x-1)^2 + 1 \). This substitution simplifies the expression.
- Don't forget to adjust the differential: Find \( du \), which is often simple when the substitution is a polynomial.
Completing the Square
Completing the square is a helpful algebraic technique that transforms quadratic expressions into a simpler, solvable form.
This is particularly useful when the integrals involve quadratic denominators.
Let's see how it is applied:
This is particularly useful when the integrals involve quadratic denominators.
Let's see how it is applied:
- Take the expression \( x^2 - 2x + 2 \).
- Try to express it in the form \((x - a)^2 + b \).
- Notice that \( x^2 - 2x + 1 \) completes the square, resulting in \((x-1)^2\).
- Add and subtract the same term to adjust: \((x-1)^2 + 1 \). This trick simplifies the quadratic.
Definite Integral Evaluation
Evaluating a definite integral involves determining the area under a curve from one point to another.
With substitution and completing the square done, it鈥檚 time to integrate.
In our example:
With substitution and completing the square done, it鈥檚 time to integrate.
In our example:
- The integral becomes \( \int_{0}^{1} \frac{8 \, dt}{t^2 + 1} \), after substitution.
- Recognize the antiderivative of \( \frac{1}{t^2 + 1} \) is \( \arctan(t) \).
- Thus, the evaluated integral becomes \( 8 \arctan(t) \).
- Apply the limits: from \( t = 0 \) to \( t = 1 \).
- Calculate the definite integral: \( 8 (\arctan(1) - \arctan(0)) \).
- Since \( \arctan(1) = \frac{\pi}{4} \) and \( \arctan(0) = 0 \), the result is \( 2\pi \).
