/*! 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 4 Let \(B(x):=-\frac{1}{2} x^{2}\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 4

Let \(B(x):=-\frac{1}{2} x^{2}\) for \(x<0\) and \(B(x):=\frac{1}{2} x^{2}\) for \(x \geq 0 .\) Show that \(\int_{a}^{b}|x| d x=B(b)-B(a)\).

Short Answer

Expert verified
The integral of the absolute function from 'a' to 'b' can be simplified to \(B(b)-B(a)\), following the properties of the absolute function and integral calculus.

Step by step solution

01

Understand the absolute value function

An absolute value function, \(|x|\), returns the 'non-negative' value of 'x', meaning when 'x' is negative we simply change the sign to positive. This is why we have the piecewise function \(B(x)\) where for negative 'x', \(B(x) = -\frac{1}{2} x^{2}\), and for 'x' greater than or equal to 0, \(B(x) = \frac{1}{2} x^{2}\). This is effectively the absolute value function.
02

Integrate |x| for the given interval

Generally, to integrate the absolute value function from 'a' to 'b', we have to consider the two cases: if 'a' and 'b' are both positive then we integrate normally; otherwise, if 'a' is negative (and 'b' is positive), we have to split this integral into two parts at '0' (due to the nature of the absolute value function), where we integrate from 'a' to '0' and '0' to 'b'. We have to take the negative integral for the negative part (as the absolute value will convert this negative area to positive).
03

Apply the Integral properties

Considering the antiderivative of \(|x|\), which is \(B(x)\), we will apply the second fundamental theorem of calculus. This theorem states that if a function is continuous on the interval [a, b] and F is an antiderivative of f on [a, b], then the definite integral from a to b of f with respect to x is equal to F(b) 鈭� F(a). Here, our 'a' can be either negative or 0, and similarly with 'b'. So the integral will directly become \(B(b) - B(a)\), which is our final simplified equation.

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

Let \(f:[a, b] \rightarrow \mathbb{R}\) and let \(C \in \mathbb{R}\). (a) If \(\Phi:[a, b] \rightarrow \mathbb{R}\) is an antiderivative of \(f\) on \([a, b]\), show that \(\Phi_{C}(x):=\Phi(x)+C\) is also an antiderivative of \(f\) on \([a, b]\). (b) If \(\Phi_{1}\) and \(\Phi_{2}\) are antiderivatives of \(f\) on \([a, b]\), show that \(\Phi_{1}-\Phi_{2}\) is a constant function on \([a, b]\).

Approximate the indicated integrals, giving estimates for the error. Use a calculator to obtain a high degree of precision.$$ \int_{0}^{2}\left(4+x^{3}\right)^{1 / 2} d x $$

Approximate the indicated integrals, giving estimates for the error. Use a calculator to obtain a high degree of precision.$$ \int_{0}^{\pi / 2} \sqrt{\sin x} d x $$

Suppose that \(f:[a, b\\} \rightarrow \mathbb{R}\) and that \(n \in \mathbb{N}\). Let \(\mathcal{P}_{n}\) be the partition of \([a, b]\) into \(n\) subintervals having equal lengths, so that \(x_{i}:=a+i(b-a) / n\) for \(i=0,1, \cdots, n\). Let \(L_{n}(f):=S\left(f ; \dot{\mathcal{P}}_{n . l}\right)\) and \(R_{n}(f):=S\left(f ; \dot{\mathcal{P}}_{n, r}\right)\), where \(\dot{\mathcal{P}}_{n, l}\) has its tags at the left endpoints, and \(\dot{\mathcal{P}}_{n, r}\) has its tags at the right endpoints of the subintervals \(\left[x_{i-1}, x_{i}\right]\). (a) If \(f\) is increasing on \([a, b]\), show that \(L_{n}(f) \leq R_{n}(f)\) and that $$ 0 \leq R_{n}(f)-L_{n}(f)=(f(b)-f(a)) \cdot \frac{(b-a)}{n} $$ (b) Show that \(f(a)(b-a) \leq L_{n}(f) \leq \int_{a}^{b} f \leq R_{n}(f) \leq f(b)(b-a)\) (c) If \(f\) is decreasing on \([a, b]\), obtain an inequality similar to that in (a).

Show that if \(f^{\prime \prime}(x) \geq 0\) on \([a, b]\) (that is, if \(f\) is convex on \(\left.[a, b]\right)\), then for any natural numbers \(m, n\) we have \(M_{n}(f) \leq \int_{a}^{b} f(x) d x \leq T_{m}(f)\). If \(f^{\prime \prime}(x) \leq 0\) on \([a, b]\), this inequality is reversed.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

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.