/*! 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 24 Use areas to evaluate the integr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 24

Use areas to evaluate the integrals in Exercises \(23-26\) $$ \int_{0}^{b} 4 x d x, \quad b > 0 $$

Short Answer

Expert verified
The integral evaluates to \( 2b^2 \).

Step by step solution

01

Identify the Area

The given problem involves evaluating the integral \( \int_{0}^{b} 4x \, dx \) using the concept of area. We recognize that the function \( f(x) = 4x \) is a straight line, which forms a triangle between \( x = 0 \) and \( x = b \). Our task is to find the area under this line between these x-values.
02

Determine the Shape and Decomposition

Since the integral represents an area under the line, and the function is linear, the region under the curve from \( x = 0 \) to \( x = b \) forms a right triangle. The base of this triangle is along the x-axis from 0 to b and rises to a height of \( 4b \) at \( x = b \).
03

Calculate the Area of the Triangle

For a triangle, the area \( A \) is calculated using the formula \( A = \frac{1}{2} \times \text{base} \times \text{height} \). Here, the base of the triangle is \( b \) and the height is \( 4b \). So, the area is:\[A = \frac{1}{2} \times b \times 4b = 2b^2\]
04

Conclude the Integral Evaluation Using Area

Since the integral from 0 to \( b \) of \( 4x \, dx \) represents the area under the line (which is the area of the triangle), the value of the integral is \( 2b^2 \).

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.

Area under a Curve
When you hear 'area under a curve,' it means the space trapped between the curve, the x-axis, and the boundaries on the left and right (here from 0 to b). Calculating this area for a linear function like \( f(x) = 4x \), can often be simplified by identifying geometric shapes. For straight lines, this usually means recognizing triangles or trapezoids. By finding the area, you can evaluate definite integrals graphically rather than solving them analytically. Integrals that measure the area are always between two points on the x-axis, providing a physical interpretation as an accumulation of space.
Triangle Area Formula
Triangles are fundamental shapes in geometry, and their area represents half the space of a corresponding rectangle. You calculate this with the formula: \( A = \frac{1}{2} \times \text{base} \times \text{height} \).
For our exercise, this formula is key. The line \( f(x) = 4x \) when graphed, forms a right triangle between the x-axis, the vertical line at x = b, and the line itself. The triangle's base is \( b \), and its height is \( 4b \), leading to an area \( 2b^2 \).
By knowing the triangle area formula, integrating linear functions becomes intuitive, as you convert an algebraic problem into a visual one.
Linear Functions
Linear functions are equations where the graph forms a straight line. They are straightforward and predictable. A typical form is \( f(x) = mx + c \), where \( m \) is the slope and \( c \) the y-intercept.
In our example, \( f(x) = 4x \), it's a line passing through the origin with a slope of 4, indicating a steep rise. This slope informs us about how fast the function value changes as x increases.
Understanding linear functions is crucial because many real-world processes can be simplified to linearity. Recognizing their graphs lets you use geometric shapes to solve integrals easily, confirming how straight lines relate to area calculations.

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

Evaluate the integrals in Exercises \(13-48\) . $$ \int \frac{1}{\theta^{2}} \sin \frac{1}{\theta} \cos \frac{1}{\theta} d \theta $$

Use the Substitution Formula in Theorem 6 to evaluate the integrals in Exercises \(1-24 .\) $$ \text { a. }\int_{0}^{1} r \sqrt{1-r^{2}} d r \quad \text { b. } \int_{-1}^{1} r \sqrt{1-r^{2}} d r $$

If \(f\) is a continuous function, find the value of the integral $$I=\int_{0}^{a} \frac{f(x) d x}{f(x)+f(a-x)}$$ by making the substitution \(u=a-x\) and adding the resulting integral to \(I .\)

It looks as if we can integrate 2 sin \(x \cos x\) with respect to \(x\) in three different ways: $$ \begin{aligned} \text { a. } \int 2 \sin x \cos x d x &=\int 2 u d u \quad u=\sin x \\ &=u^{2}+C_{1}=\sin ^{2} x+C_{1} \\ \text { b. } \int 2 \sin x \cos x d x &=\int-2 u d u \quad u=\cos x \\ &=-u^{2}+C_{2}=-\cos ^{2} x+C_{2} \\\ \text { c. } \int 2 \sin x \cos x d x &=\int \sin 2 x d x \quad 2 \sin x \cos x=\sin 2 x \\ &=-\frac{\cos 2 x}{2}+C_{3} \end{aligned} $$ Can all three integrations be correct? Give reasons for your answer.

Evaluate the integrals in Exercises \(13-48\) . $$ \int 3 x^{5} \sqrt{x^{3}+1} d x $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

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.