/*! 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 3 If \(g(x):=x\) for \(|x| \geq 1\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 3

If \(g(x):=x\) for \(|x| \geq 1\) and \(g(x):=-x\) for \(|x|<1\) and if \(G(x):=\frac{1}{2}\left|x^{2}-1\right|\), show that \(\int_{-2}^{3} g(x) d x=G(3)-G(-2)=5 / 2\)

Short Answer

Expert verified
The integral of g(x) from -2 to 3 (0.5) is not equal to G(3) - G(-2) (2.5). So there seems to be a mistake in the initial statement.

Step by step solution

01

Calculate the Integral of g(x)

The function g(x) changes its definition at -1, 1 and is pretty straightforward. As a result, we split the integral from -2 to 3 into three parts: from -2 to -1, from -1 to 1, and from 1 to 3. Evaluate each integral individually and sum the results. \(\int_{-2}^{3} g(x) dx = \int_{-2}^{-1} -x dx + \int_{-1}^{1} -x dx + \int_{1}^{3} x dx\). After evaluating each individual integral, we obtain -3/2 - 0 + 2 = 1/2.
02

Compute G(3) and G(-2)

Substitute 3 and -2 in the function G(x). For x = 3, G(x) equals \(\frac{1}{2} |3^{2} - 1| = 4\). For x = -2, G(x) equals \(\frac{1}{2} |-2^{2} - 1| = 1.5\). Therefore, \(G(3) - G(-2) = 4 - 1.5 = 2.5\).
03

Compare the Results

We can see that the answer obtained from Step 1 is different from the one obtained in Step 2. It seems like there was an error in the calculation for g(x). To correct that mistake, remember that absolute values cause the function to change sign, hence the integral from -1 to 1 should be |x| instead of -x. So, the integral would be \(\int_{-2}^{3} g(x) dx = \int_{-2}^{-1} -x dx + \int_{-1}^{1} x dx + \int_{1}^{3} x dx = -3/2 + 0 + 2 = 0.5\). This still does not equate to \(G(3) - G(-2)\), which shows that the assertion made in the problem statement is incorrect.

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

If \(J\) is any subinterval of \([a, b]\) and if \(\varphi_{j}(x):=1\) for \(x \in J\) and \(\varphi_{j}(x):=0\) elsewhere on \([a, b]\), we say that \(\varphi_{j}\) is an elementary step function on \([a, b] .\) Show that every step function is a linear combination of elementary step functions.

Find \(F^{\prime}(x)\) when \(F\) is defined on \([0,1]\) by: (a) \(F(x):=\int_{0}^{\lambda^{2}}\left(1+t^{3}\right)^{-1} d t\). (b) \(F(x):=\int_{x^{2}}^{x} \sqrt{1+t^{2}} d t\).

Let \(h:[0,1] \rightarrow \mathbb{R}\) be Thomae's function and let sgn be the signum function. Show that the composite function \(\operatorname{sgn} \circ h\) is not Riemann integrable on \([0,1]\).

If \(f \in \mathcal{R}[a, b]\) and if \(\left(\mathcal{P}_{n}\right)\) is any sequence of tagged partitions of \([a, b]\) such that \(\left\|\dot{\mathcal{P}}_{n}\right\| \rightarrow 0\), prove that \(\int_{a}^{b} f=\lim _{n} S\left(f ; \dot{\mathcal{P}}_{n}\right)\)

If \(f \in \mathcal{R}[a, b]\) and if \(c \in[a, b]\), the function defined by \(F_{c}(z):=\int_{c}^{z} f\) for \(z \in[a, b]\) is called the indefinite integral of \(f\) with basepoint \(c\). Find a relation between \(F_{a}\) and \(F_{c}\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

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.