/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 12

Evaluate the following integrals. $$\int s e^{-2 s} d s$$

Short Answer

Expert verified
Question: Evaluate the integral 鈭玸e^(-2s)ds. Answer: 鈭玸e^(-2s)ds = -1/2 * s * e^(-2s) + 1/4 * e^(-2s) + K (where K is a constant).

Step by step solution

01

Apply integration by parts formula

Using integration by parts, let: u = s, dv = e^{-2s} ds Now we need to find du (the derivative of u) and v (the antiderivative of dv): du = ds, v = \int e^{-2s} ds
02

Find the antiderivative of v

To find v, we need to evaluate the integral: v = \int e^{-2s} ds Using substitution method, let p = -2s dp = -2 ds Hence, ds = (-1/2) dp Substitute 'p' and 'ds' into the integral equation: v = \int e^{p} (-1/2) dp Now, integrate the exponential term: v = (-1/2) \int e^{p} dp v = (-1/2) e^{p} + C Substitute back to 's': v = (-1/2) e^{-2s} + C
03

Apply the integration by parts formula

Applying the integration by parts formula: \int s e^{-2 s} d s = uv - \int v du Substitute the values of u, v, and du: \int s e^{-2 s} d s = s\left( - \frac{1}{2} e^{-2s} + C\right) - \int (-\frac{1}{2} e^{-2s} + C) ds
04

Integrate v du

Now, we need to evaluate the integral: \int (-\frac{1}{2} e^{-2s} + C) ds =-\frac{1}{2} \int e^{-2s} ds + C \int ds We can use the earlier substitution method to evaluate the integral of the exponential part: =-\frac{1}{2} ((-\frac{1}{2}) e^{-2s} + C) + C s + D =(-\frac{1}{4}) e^{-2s} + \frac{1}{2} Cs + Cs + D (where D is another constant)
05

Substitute back the values

Now, substitute the values back into the integration by parts equation: \int s e^{-2 s} d s = s\left( - \frac{1}{2} e^{-2s} + C\right) - [(-\frac{1}{4}) e^{-2s} + \frac{1}{2} Cs + Cs + D] = -\frac{1}{2}s e^{-2s} + Cs^2 + \frac{1}{4} e^{-2s} - \frac{1}{2} Cs^2 - Cs^2 - D Combine the like terms to get the final solution: \int s e^{-2 s} d s = -\frac{1}{2}s e^{-2 s} + \frac{1}{4} e^{-2 s} + K where K = D (a constant).

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove that the Trapezoid Rule is exact (no error) when approximating the definite integral of a linear function.

Refer to Theorem 2 and let \(f(x)=\sin e^{x}\) a. Find a Trapezoid Rule approximation to \(\int_{0}^{1} \sin \left(e^{x}\right) d x\) using \(n=40\) subintervals. b. Calculate \(f^{\prime \prime}(x)\) c. Explain why \(\left|f^{\prime \prime}(x)\right|<6\) on \([0,1],\) given that \(e<3\). (Hint: Graph \(\left.f^{\prime \prime} .\right)\) d. Find an upper bound on the absolute error in the estimate found in part (a) using Theorem 2.

Consider the family of functions \(f(x)=1 / x^{p},\) where \(p\) is a real number. For what values of \(p\) does the integral \(\int_{0}^{1} f(x) d x\) exist? What is its value?

Use integration by parts to evaluate the following integrals. $$\int_{0}^{\infty} x e^{-x} d x$$

Bob and Bruce bake bagels (shaped like tori). They both make standard bagels that have an inner radius of 0.5 in and an outer radius of 2.5 in. Bob plans to increase the volume of his bagels by decreasing the inner radius by \(20 \%\) (leaving the outer radius unchanged). Bruce plans to increase the volume of his bagels by increasing the outer radius by \(20 \%\) (leaving the inner radius unchanged). Whose new bagels will have the greater volume? Does this result depend on the size of the original bagels? Explain.
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.