Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 16
The integrals in Exercises \(1-34\) converge. Evaluate the integrals without using tables. $$\int_{0}^{2} \frac{s+1}{\sqrt{4-s^{2}}} d s$$
Short Answer
Expert verified
The integral evaluates to \( 2 + \frac{\pi}{2} \).
Step by step solution
01
Rewrite the Integral
Our integral is \( \int_{0}^{2} \frac{s+1}{\sqrt{4-s^{2}}} \, ds \). Notice that we can break this integral into two separate integrals: \( \int_{0}^{2} \frac{s}{\sqrt{4-s^{2}}} \, ds \) and \( \int_{0}^{2} \frac{1}{\sqrt{4-s^{2}}} \, ds \). This is because the integrand \( \frac{s+1}{\sqrt{4-s^{2}}} \) can be split as \( \frac{s}{\sqrt{4-s^{2}}} + \frac{1}{\sqrt{4-s^{2}}} \).
02
Evaluate the First Integral
Consider the first integral: \( \int_{0}^{2} \frac{s}{\sqrt{4-s^{2}}} \, ds \).Use substitution: Let \( u = 4 - s^2 \), then \( du = -2s \, ds \) or \( s \, ds = -\frac{1}{2} du \). When \( s = 0 \), \( u = 4 \). When \( s = 2 \), \( u = 0 \).This changes the integral to:\[ -\frac{1}{2} \int_{4}^{0} \frac{1}{\sqrt{u}} \, du. \]The integral \( \int \frac{1}{\sqrt{u}} \, du \) is \( 2\sqrt{u} \). Thus, the evaluated integral is:\[ -\frac{1}{2}(2\sqrt{u}) \Bigg|_{4}^{0} = -\sqrt{u} \Bigg|_{4}^{0} = 0 - (-2) = 2. \]
03
Evaluate the Second Integral
Now consider the second integral: \( \int_{0}^{2} \frac{1}{\sqrt{4-s^{2}}} \, ds \).This is a standard integral that is derived from the arcsine function. It is of the form \( \arcsin\left( \frac{s}{2} \right) \) after considering the constant terms. Evaluating from \( 0 \) to \( 2 \):\[ \arcsin\left( \frac{s}{2} \right) \Bigg|_{0}^{2} = \arcsin(1) - \arcsin(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \]
04
Combine the Integrals
Finally, sum the results from the first and second integrals:\( 2 + \frac{\pi}{2} = 2 + \frac{\pi}{2}. \)Thus, the value of the original integral is \( 2 + \frac{\pi}{2} \).
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 useful tool for solving integrals that are not straightforward. It involves changing variables to simplify the integral into a form that is easier to handle. In essence, substitution is like "rebranding" the integral with a new variable that makes the solution process more convenient.
Here's how you can employ the substitution method:
Here's how you can employ the substitution method:
- Identify a part of the integrand that complicates the evaluation process. In this exercise, it was the expression under the square root, \(4 - s^2\).
- Set your new variable \(u\) as this complex expression, such as \(u = 4 - s^2\). Consequently, find the differential by differentiating: \(du = -2s \, ds\).
- Substitute this new expression \(u\) into the integral in place of \(4 - s^2\) and \(s \, ds\).
- Remember to change the limits of integration according to the new variable \(u\). When \(s = 0\), \(u = 4\), and when \(s = 2\), \(u = 0\).
- Evaluate the transformed integral, which should now be simpler, and switch back to the original variable if needed before integrating the final result.
Arcsine Function
The arcsine function is an inverse trigonometric function denoted by \(\arcsin(x)\). It gives the angle whose sine is \(x\). In integral calculus, the arcsine function often appears when dealing with integrals involving square roots of the form \(\sqrt{a^2 - x^2}\).
Here is how the arcsine function relates to this exercise:
Here is how the arcsine function relates to this exercise:
- The integral \(\int \frac{1}{\sqrt{4-s^2}} \, ds\) directly matches the form required to apply the arcsine function. Here, \(a = 2\).
- The integral yields \(\arcsin\left(\frac{s}{2}\right)\) with respect to \(s\).
- To evaluate the definite integral, substitute the upper and lower limits into the arcsine result. In this case, from \(0\) to \(2\), giving \(\arcsin\left(1\right) - \arcsin\left(0\right) = \frac{\pi}{2} - 0 = \frac{\pi}{2}\).
Integral Evaluation
Integral evaluation is the process of finding the value of an integral. It encompasses several techniques, including substitution and recognizing standard forms like those involving inverse trigonometric functions.
To effectively evaluate an integral:
To effectively evaluate an integral:
- Break the integrand into simpler components if necessary, as seen with splitting \(\frac{s+1}{\sqrt{4-s^2}}\) into two separate integrals.
- Use substitution methods to handle integrals that involve complex expressions, converting them into easier forms.
- Recognize standard integral forms, such as those involving the arcsine function, to apply relevant formulas quickly.
- Calculate each integral individually and sum them to find the total value. In this exercise, \(2 + \frac{\pi}{2}\) represents the combined value of both evaluated integrals.
