/*! 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 Write the given (total) area as ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 12

Write the given (total) area as an integral or sum of integrals. The area above the \(x\) -axis and below \(y=4 x-x^{2}\).

Short Answer

Expert verified
The total area under the curve \(y=4x-x^{2}\) from \(x = 0\) to \(x = 4\) can be expressed as the integral \(\int_{0}^{4} (4x-x^{2}) dx\).

Step by step solution

01

Identify the limits of the integral

The bounding points are the values of \(x\) where the given equation equals 0, which are \[0 = 4x-x^{2} \Rightarrow x (4 - x) = 0 \]. This yields two solutions: \(x = 0\) and \(x = 4\). These are the limits of the integral.
02

Writing the integral

The area under the curve can be found by taking the integral of the curve from the lower limit \(x = 0\) to the upper limit \(x = 4\). So it is to be written as follows: \[\int_{0}^{4} (4x-x^{2}) dx\].
03

Calculating the integral

Gather the constants out of the integral and then integrate each term to find its primitive function: \[\int_{0}^{4} (4x-x^{2}) dx = \left[2x^{2} - \frac{1}{3}x^{3}\right]_{0}^{4}\].

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.

Finding Limits of Integration
When dealing with definite integrals, the first step is to identify the limits of integration. These boundaries determine where on the x-axis our integration, thereby the area calculation, starts and ends.
To find these limits, especially when given a curve, set the function equal to zero. The points where it intersects the x-axis form the lower and upper limits of integration.
  • For instance, with the curve described by the function \( y = 4x - x^2 \), setting \( y = 0 \) gives us the equation \( 4x - x^2 = 0 \).
  • Solving \( x(4 - x) = 0 \) reveals the solutions \( x = 0 \) and \( x = 4 \), our integration limits.
These points show the sections over which we will calculate the area under the curve. They are critical as they ensure we only integrate over the specific section in question.
Area Under a Curve
Finding the area between a curve and the x-axis involves integrating the function within its limits. This process helps in various applications, such as calculating distances, volumes, or net changes.
  • The integral essentially sums up the infinitely small slices of area between the curve and the x-axis over a specific interval.
  • In our case, we calculate the integral from the leftmost point \( x = 0 \) to the rightmost point \( x = 4 \).
By writing the integral as \( \int_{0}^{4} (4x - x^2) \ dx \), we represent the area under the curve between these two x-values. It is crucial to interpret the function correctly since we are looking for the area above the x-axis, ensuring that all values are positive over the interval.
Polynomial Integration
Integrating polynomials involves finding the anti-derivative, which gives us a new function that represents accumulated area. This is usually straightforward, provided we follow some basic rules.
  • Each term in the polynomial is handled separately: increase the exponent by one and divide by the new exponent.
  • For the integral of \( 4x \), we increase the exponent to 2, giving us \( 2x^2 \).
  • For \( -x^2 \), we apply the same approach to get \(-\frac{1}{3}x^3\).
After integrating, we find the area by substituting back the limits of integration. Plug these limits into the integrated function and subtract: \[\left[2x^{2} - \frac{1}{3}x^{3}\right]_{0}^{4} = \left(2(4)^2 - \frac{1}{3}(4)^3\right) - \left(2(0)^2 - \frac{1}{3}(0)^3\right) \]. This calculation gives us the precise value of the area under the curve within the specified bounds.

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

Identify the integrals to which the Fundamental Theorem of Calculus applies; the other integrals are called improper integrals. (a) \(\int_{0}^{1} \frac{1}{\sqrt{x+2}} d x \quad\) (b) \(\int_{0}^{2} \frac{1}{(x-3)^{2}} d x \quad\) (c) \(\int_{0}^{2} \sec x d x\)

The data come from a pneumotachograph, which measures air flow through the throat (in liters per second). The integral of the air flow equals the volume of air exhaled. Estimate this volume. $$\begin{array}{|l|l|l|l|l|l|l|l|} \hline t(\mathrm{s}) & 0 & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 & 1.2 \\ \hline f(t)(1 / s) & 0 & 0.2 & 0.4 & 1.0 & 1.6 & 2.0 & 2.2 \\ \hline \end{array}$$ $$\begin{array}{|l|l|l|l|l|l|l|} \hline t(\mathrm{s}) & 1.4 & 1.6 & 1.8 & 2.0 & 2.2 & 2.4 \\ \hline f(t)(1 / \mathrm{s}) & 2.0 & 1.6 & 1.2 & 0.6 & 0.2 & 0 \\ \hline \end{array}$$

A telegraph cable is made of an outer winding around an inner core. If \(x\) is defined as the core radius divided by the outer radius, the transmission speed is proportional to \(s(x)=x^{2} \ln (1 / x) .\) Find an \(x\) that maximizes the transmission speed.

The table shows the temperature at different times of the day. Estimate the average temperature using (a) right-endpoint evaluation and (b) left-endpoint evaluation. Explain why the estimates are different. $$\begin{array}{|l|l|l|l|l|l|l|l|l|l|} \hline \text { time } & 12: 00 & 3: 00 & 6: 00 & 9: 00 & 12: 00 & 3: 00 & 6: 00 & 9: 00 & 12: 00 \\ \hline \text { temperature } & 46 & 44 & 52 & 70 & 82 & 86 & 80 & 72 & 56 \\ \hline \end{array}$$

The impulse-momentum equation states the relationship between a force \(F(t)\) applied to an object of mass \(m\) and the resulting change in velocity \(\Delta v\) of the object. The equation is \(m \Delta v=\int_{a}^{b} F(t) d t,\) where \(\Delta v=v(b)-v(a) .\) Suppose that the force of a baseball bat on a ball is approximately \(F(t)=9-10^{8}(t-0.0003)^{2}\) thousand pounds, for \(t\) between 0 and 0.0006 second. What is the maximum force on the ball? Using \(m=0.01\) for the mass of a baseball, estimate the change in velocity \(\Delta v\) (in \(\mathrm{f} t / \mathrm{s}\) ).
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

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.