Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 66
Completing the square Evaluate the following integrals. $$\int_{1}^{4} \frac{d t}{t^{2}-2 t+10}$$
Short Answer
Expert verified
Question: Evaluate the given integral: $$\int_{1}^{4} \frac{d t}{t^{2}-2 t+10}$$
Answer: $$\frac{\pi}{6}$$
Step by step solution
01
Complete the square
First, we need to complete the square for the quadratic expression in the denominator. The expression is \(t^2 - 2t + 10\). To complete the square, take half of the coefficient of the linear term (-2), square it, and add and subtract that value to the expression.
$$t^2 - 2t + 10 = (t^2 - 2t + 1) + 9 = (t - 1)^2 + 9$$
02
Rewrite the integral
Now, we substitute the completed square form of the denominator back into the integral:
$$\int_{1}^{4} \frac{d t}{(t - 1)^2 + 9}$$
03
Apply the trigonometric substitution
We can perform the trigonometric substitution by setting:
$$t - 1 = 3\sin(\theta) \Rightarrow dt = 3\cos(\theta)d\theta$$
Now, replace \(t\) and \(dt\) with the trigonometric substitution:
$$\int_{\theta_1}^{\theta_4} \frac{3\cos(\theta)}{(3\sin(\theta))^2 + 9}d\theta$$
Where \(\theta_1 = \sin^{-1}\left(\frac{1-1}{3}\right) = 0\) and \(\theta_4 = \sin^{-1}\left(\frac{4-1}{3}\right)\).
04
Simplify the integral expression
Simplify the obtained integral expression:
$$\int_{0}^{\sin^{-1}(1)} \frac{3\cos(\theta)}{9\sin^2(\theta) + 9}d\theta = \int_{0}^{\sin^{-1}(1)} \frac{\cos(\theta)}{3\sin^2(\theta) + 3}d\theta$$
$$=\frac{1}{3}\int_{0}^{\sin^{-1}(1)} \frac{\cos(\theta)}{\sin^2(\theta) + 1}d\theta = \frac{1}{3}\int_{0}^{\sin^{-1}(1)} d\theta$$
By substituting back \(\theta\) to \(t\) (Inverse trig substitution):
$$\frac{1}{3}\int_{0}^{\sin^{-1}(1)} d\theta = \frac{1}{3}[\theta]_{0}^{\sin^{-1}(1)} = \frac{1}{3}\left[\sin^{-1}(1)-0\right] = \frac{1}{3}\left[\frac{\pi}{2}\right]$$
05
Evaluate the integral
Finally, evaluate the integral:
$$\frac{1}{3}\left[\frac{\pi}{2}\right] = \frac{\pi}{6}$$
Thus, the value of the integral is:
$$\int_{1}^{4} \frac{d t}{t^{2}-2 t+10} = \frac{\pi}{6}$$
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.
Completing the Square
Completing the square is a powerful algebraic technique used to transform a quadratic expression into a perfect square trinomial plus or minus a constant. This method not only aids in solving quadratic equations but also plays a significant role in calculus, especially in integration problems involving quadratic expressions.
To complete the square for a quadratic expression like \(t^2 - 2t + 10\), here鈥檚 what you do:
Applying these steps to our expression, we have:
\[t^2 - 2t + 10 = (t - 1)^2 + 9\]
By converting the expression into this form, the integral becomes easier to handle, often allowing for trigonometric substitution.
To complete the square for a quadratic expression like \(t^2 - 2t + 10\), here鈥檚 what you do:
- Identify the coefficient of the linear term (here it's \(-2\)).
- Take half of that coefficient, giving you \(-1\).
- Square that result, resulting in \(1\).
- Add and subtract this square inside the expression.
Applying these steps to our expression, we have:
\[t^2 - 2t + 10 = (t - 1)^2 + 9\]
By converting the expression into this form, the integral becomes easier to handle, often allowing for trigonometric substitution.
Trigonometric Substitution
Trigonometric substitution is a method used in calculus to simplify integrals, especially those involving square roots or quadratic expressions. It leverages trigonometric identities for easier evaluation by changing variables.
For our integral:
- After completing the square, the denominator becomes \((t - 1)^2 + 9\), resembling the identity \(a^2 + x^2\).
- We use the substitution \(t - 1 = 3 \sin(\theta)\), so that the expression matches the identity. Here, \(3 heta\) aligns with \(a\cdot\sin(\theta)\).
- Differentiating, \(dt = 3\cos(\theta)d\theta\).
Substituting these into the integral, we can simplify:\[\int \frac{3 \cos(\theta)}{(3 \sin(\theta))^2 + 9}d\theta\]
With trigonometric identities, \(\sin^2(\theta) + \cos^2(\theta) = 1\), it simplifies to an easily integrable form.
For our integral:
- After completing the square, the denominator becomes \((t - 1)^2 + 9\), resembling the identity \(a^2 + x^2\).
- We use the substitution \(t - 1 = 3 \sin(\theta)\), so that the expression matches the identity. Here, \(3 heta\) aligns with \(a\cdot\sin(\theta)\).
- Differentiating, \(dt = 3\cos(\theta)d\theta\).
Substituting these into the integral, we can simplify:\[\int \frac{3 \cos(\theta)}{(3 \sin(\theta))^2 + 9}d\theta\]
With trigonometric identities, \(\sin^2(\theta) + \cos^2(\theta) = 1\), it simplifies to an easily integrable form.
Definite Integrals
Definite integrals are a staple in calculus, representing the signed area under a curve between two points. Unlike indefinite integrals, which have a constant of integration, definite integrals result in a real number.
They follow the form:
\[\int_a^b f(x)dx\]
Where \(a\) and \(b\) are the limits of integration.
In our exercise, the integral \(\int_{1}^{4} \frac{dt}{t^{2} - 2t + 10}\) has bounds from 1 to 4. Using our trigonometric substitution and simplifying the expression, we're then able to evaluate the integral from \(\theta = 0\) to \(\theta = \sin^{-1}(1)\), translating into exact limits in the \(\theta\) domain.
After calculating, we obtain the evaluated definite integral as \(\frac{\pi}{6}\), representing the exact area under the curve from \(t=1\) to \(t=4\). This process showcases the beauty and utility of definite integrals in finding precise areas under complex functions.
They follow the form:
\[\int_a^b f(x)dx\]
Where \(a\) and \(b\) are the limits of integration.
In our exercise, the integral \(\int_{1}^{4} \frac{dt}{t^{2} - 2t + 10}\) has bounds from 1 to 4. Using our trigonometric substitution and simplifying the expression, we're then able to evaluate the integral from \(\theta = 0\) to \(\theta = \sin^{-1}(1)\), translating into exact limits in the \(\theta\) domain.
After calculating, we obtain the evaluated definite integral as \(\frac{\pi}{6}\), representing the exact area under the curve from \(t=1\) to \(t=4\). This process showcases the beauty and utility of definite integrals in finding precise areas under complex functions.
