Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 1019

\( \log 3 \int_{\log (1 / 3)} 2^{(\mathrm{x}) 2} \cdot \mathrm{x}^{3} \mathrm{~d} \mathrm{x}\) is equal to...... (a) 0 (b) \(\log 3\) (c) \(-\log 3\) (d) \(\log 2\)

Short Answer

Expert verified
-\log 3

Step by step solution

01

Evaluate the integral

To evaluate the integral, let \(u = x^2\), then the differential du is given by \(\mathrm{d}u = 2x \mathrm{~d} x\). The given integral becomes: \(\int_{\log (1 / 3)}^\infty 2^u \cdot x^3 \mathrm{~d} x =\frac{1}{2} \int_{u(\log (1 / 3))}^\infty 2^u \cdot u \mathrm{~d} u \)
02

Apply integration by parts

Integration by parts states that the integral of the product of two functions is given by: \[\int u \mathrm{~d} v = uv - \int v \mathrm{~d} u\] In our case, let \(v = 2^u\) and \(\mathrm{d}u = u \mathrm{~d}u\). Now, we need to find their counterparts: \(\mathrm{d}v = \ln(2) \cdot 2^u \mathrm{~d}u\) \(u = u\) Now, we apply the integration by parts formula: \[\frac{1}{2}\left(u \cdot 2^u - \int \ln(2) \cdot 2^u \mathrm{~d}u \right) \Big|_{u(\log (1 / 3))}^\infty\]
03

Calculate the remaining integral

Now we need to find \[\int \ln(2) \cdot 2^u \mathrm{~d}u\] Using the substitution \(t = 2^u\), thus \(\mathrm{d}t = \ln (2) \cdot 2^u \mathrm{~d}u\): \[\int \ln(2) \cdot 2^u \mathrm{~d}u = \int \mathrm{d}t = t\] Therefore, substituting back, we have: \[\int \ln(2) \cdot 2^u \mathrm{~d}u = 2^u\]
04

Combine the results

Replace the integral with the result found in Step 3: \[\frac{1}{2}\left(u \cdot 2^u - 2^u\right) \Big|_{u(\log (1 / 3))}^\infty\] Now, we need to calculate this expression at the limits: \[\frac{1}{2}(\infty \cdot 2^{\infty} - 2^{\infty}) - \frac{1}{2}(0 \cdot 2^{0} - 2^{0}) = \frac{1}{2}(0 - 1) = -\frac{1}{2} \]
05

Multiply by log3 and compare options

Now, we multiply the result by \(\log 3\): \(\log 3 \int_{\log (1 / 3)}^{\infty} 2^{x^2} \cdot x^3 \mathrm{~d} x = -\frac{1}{2} \log 3\) Comparing the options, we see that the correct answer is (c) \(-\log 3\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

The parabolas \(\mathrm{y}^{2}=4 \mathrm{x}\) and \(\mathrm{x}^{2}=4 \mathrm{y}\) divide the square region bounded the lines \(x=4, y=4\) and the coordinate axes. If \(S_{1}\), \(\mathrm{S}_{2}, \mathrm{~S}_{3}\) are respectively the areas of these parts numbered from top to bottom then \(S_{1}: S_{2}: S_{3}\) is........ (a) \(1: 2: 3\) (b) \(2: 1: 2\) (c) \(3: 2: 3\) (d) \(1: 1: 1\)

\(1 \int_{0} 3 \sqrt{\left(x^{3}-x^{4}\right) d x}\) is equal to....... (a) \((1 / 2)\) (b) \((3 / 7)\) (c) \((9 / 28)\) (d) \((29 / 28)\)

The area of the region bounded by the lines \(y=m x, x=1\), \(\mathrm{x}=2\) and \(\mathrm{x}\) -axis is \(6 \mathrm{Sq}\). unit then \(\mathrm{m}\) is....... (a) 1 (b) 2 (c) 3 (d) 4

The area bounded by the curves \(x^{2}=y\) and \(2 x+y-8=0\) and \(y\) -axis in the second quadrant is...... (a) 9 Sq. unit (b) 18 Sq. unit (c) \((80 / 3)\) Sq. unit (d) 36 Sq. unit

\(1\mid={ }^{1} \int_{-1}\left(x^{7}+\cos ^{-1} x\right) d x\) then \(\cos \mid\) is equal to....... (a) 1 (b) 0 (c) \(-1\) (d) \((1 / 2)\)
See all solutions

Recommended explanations on English Textbooks

Essay Prompts

Read Explanation

Linguistic Terms

Read Explanation

Listening and Speaking

Read Explanation

Phonetics

Read Explanation

Global English

Read Explanation

Multiple Choice Questions

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.

Sign-up for free