Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 7: Problem 48
Evaluate. $$\int_{4}^{6} \frac{d x}{(x-3) \sqrt{x^{2}-6 x+8}}$$.
Short Answer
Expert verified
The solution to the integral \(\int_{4}^{6} \frac{d x}{(x-3) \sqrt{x^{2}-6 x+8}}\) is \(\frac{\sqrt{3}\pi}{2}\).
Step by step solution
01
Factor the quadratic term under the square root
We first simplify the quadratic term under the square root:
\[
x^2 - 6x + 8 = (x - 4)(x - 2).
\]
Now our integral becomes
\[
\int_{4}^{6} \frac{dx}{(x-3) \sqrt{(x - 4)(x - 2)}}.
\]
02
Make the substitution \(x - 4 = u^2\)
Let
\[
x - 4 = u^2.
\]
Then
\[
dx = 2u\, du.
\]
When \(x = 4\), \(u^2 = 4 - 4 = 0\), so \(u = 0\). When \(x = 6\), \(u^2 = 6 - 4 = 2\), so \(u = \sqrt{2}\).
Now we change the integral to the u variable:
\begin{align*}
\int_{4}^{6} \frac{dx}{(x-3) \sqrt{(x - 4)(x - 2)}} &= \int_{0}^{\sqrt{2}} \frac{2u\, du}{\left((u^2 + 1)(u^2 + 2)\right)^{1/2}} \\
&= 2\int_{0}^{\sqrt{2}} \frac{u}{\sqrt{u^2(u^2 + 3)}\, } du.
\end{align*}
03
Make the substitution \(u = \sqrt{3}\sin{\theta}\)
Now, we will make a trigonometric substitution to simplify the expression under the square root. Let's use
\[
u = \sqrt{3} \sin{\theta}.
\]
Then
\[
du = \sqrt{3}\cos{\theta}\, d\theta.
\]
When \(u = 0\), \(\theta = 0\). When \(u = \sqrt{2}\), \(\theta = \pi/4\).
Now we change the integral to the \(\theta\) variable:
\begin{align*}
2\int_{0}^{\sqrt{2}} \frac{u}{\sqrt{u^2(u^2 + 3)}\, } du &= 2 \int_{0}^{\pi/4} \frac{\sqrt{3}\sin{\theta}}{\sqrt{3}\sin{\theta}\cos{\theta}} \sqrt{3}\cos{\theta}\, d\theta \\
&= 2\sqrt{3} \int_{0}^{\pi/4} d\theta.
\end{align*}
04
Evaluate the integral
Now we can directly evaluate the integral:
\begin{align*}
2\sqrt{3} \int_{0}^{\pi/4} d\theta &= 2\sqrt{3} \left[\theta\right]_{0}^{\pi/4} \\
&= 2\sqrt{3} \cdot \frac{\pi}{4} \\
&= \frac{\sqrt{3}\pi}{2}.
\end{align*}
So, the integral evaluates to
\[
\int_{4}^{6} \frac{dx}{(x-3) \sqrt{x^{2}-6 x+8}} = \frac{\sqrt{3}\pi}{2}.
\]
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.
Substitution Method
The Substitution Method is a key technique in integral calculus. It transforms a complex integral into a simpler one, making it easier to solve. The idea is to substitute a part of the integral with a new variable. This substitution simplifies the original integrand.
For example, in the exercise above, the original integral:
For example, in the exercise above, the original integral:
- \( \int \frac{d x}{(x-3) \sqrt{(x - 4)(x - 2)}} \)
- \(dx = 2u\, du\)
- The limits of integration change from \(x = 4\) to \(u = 0\) and from \(x = 6\) to \(u = \sqrt{2}\)
Trigonometric Substitution
Trigonometric Substitution is another powerful tool in integral calculus for integrals involving square roots of quadratic expressions. It involves substituting a trigonometric function to further simplify the integral.
In our example, the expression was modified to \(u = \sqrt{3} \sin\theta\). This clever substitution exploits the property of sine and cosine working well with square roots:
In our example, the expression was modified to \(u = \sqrt{3} \sin\theta\). This clever substitution exploits the property of sine and cosine working well with square roots:
- \(du = \sqrt{3} \cos\theta\, d\theta\)
- The integral's bounds change from \(u = 0\) (which gives \(\theta = 0\)) to \(u = \sqrt{2}\) (which gives \(\theta = \pi/4\))
Definite Integral
A Definite Integral computes the net area under a curve between two points. In this context, it transforms a problem into a finite computation rather than finding an indefinite antiderivative.
In the exercise, we computed \( \int_{4}^{6} \frac{d x}{(x-3) \sqrt{x^{2}-6 x+8}} \), which involves calculating the area from \(x=4\) to \(x=6\). By performing the substitution and adjustment of limits, the definite nature ensures that once the integration is completed via the substitution and parameter adjustment, it delivers a numeric answer.
This definite nature allows us to adjust the integration limits based on our substitutions, as seen in transitioning from \(u\) to \(\theta\) variables, ultimately yielding a concrete solution: \(\frac{\sqrt{3}\pi}{2}\).
In the exercise, we computed \( \int_{4}^{6} \frac{d x}{(x-3) \sqrt{x^{2}-6 x+8}} \), which involves calculating the area from \(x=4\) to \(x=6\). By performing the substitution and adjustment of limits, the definite nature ensures that once the integration is completed via the substitution and parameter adjustment, it delivers a numeric answer.
This definite nature allows us to adjust the integration limits based on our substitutions, as seen in transitioning from \(u\) to \(\theta\) variables, ultimately yielding a concrete solution: \(\frac{\sqrt{3}\pi}{2}\).
Quadratic Expressions
Quadratic Expressions frequently appear in calculus when solving integrals. Recognizing and factoring these expressions can simplify integration tasks.
In this specific exercise, the quadratic \(x^2 - 6x + 8\) was factored as \((x - 4)(x - 2)\). This factorization is pivotal as it allows for substitutions that simplify the expression inside the square root.
Quadratic expressions can often be simplified using:
In this specific exercise, the quadratic \(x^2 - 6x + 8\) was factored as \((x - 4)(x - 2)\). This factorization is pivotal as it allows for substitutions that simplify the expression inside the square root.
Quadratic expressions can often be simplified using:
- Completing the square
- Using the quadratic formula
- Recognizing standard factorable forms
