/*! 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 Make a sketch to show a case in ... [FREE SOLUTION] | 91Ó°ÊÓ

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Make a sketch to show a case in which the area bounded by two curves is most easily found by integrating with respect to \(x\)

Short Answer

Expert verified
Answer: The area bounded by the curves \(y = x^2\) and \(y = x\) is \(\frac{1}{6}\) square units.

Step by step solution

01

Choose two curves

Let's choose two simple curves for easy integration with respect to \(x\). We take the curves \(y = x^2\) and \(y = x\). These equations represent a quadratic function and a linear function, respectively.
02

Sketch the two curves

Draw a graph with \(x\) and \(y\) axes. Then plot the two curves \(y = x^2\) and \(y = x\) on the graph. Observe that the quadratic curve opens upwards like a parabola, while the linear curve is a straight line with a slope of \(1\) and intersects the parabola.
03

Find the intersection points of the two curves

To determine the intersection points, set the equations equal to each other: \(x^2 = x\) To solve for \(x\), first move all terms to the left-hand side: \(x^2 - x = 0\) Factor out \(x\): \(x(x - 1) = 0\) This equation has the solutions \(x = 0\) and \(x = 1\). Plug these values back into either equation to find the corresponding \(y\) values: \(y = 0^2 =0\) and \(y = 1^2 = 1\) So, the intersection points are \((0,0)\) and \((1,1)\).
04

Calculate the area bounded by the curves

To find the area, we'll integrate the difference between the two curves with respect to \(x\) over the interval from \(0\) to \(1\): \(\int_{0}^{1} (x - x^2) dx\)
05

Evaluate the integral

Apply the power rule to integration: \(\int (x - x^2) dx = \frac{x^2}{2} - \frac{x^3}{3} + C\) Now, evaluate the definite integral from \(0\) to \(1\): \(\left[\frac{1^2}{2} - \frac{1^3}{3}\right] - \left[\frac{0^2}{2} - \frac{0^3}{3}\right] = \left[\frac{1}{2} - \frac{1}{3}\right]= \frac{1}{6}\)
06

Interpret the result

The area bounded by the curves \(y = x^2\) and \(y = x\) is \(\frac{1}{6}\) square units.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area Between Curves
Understanding the concept of the area between curves can make integration more intuitive. Imagine two functions plotted on a graph. The area between these functions is essentially the region enclosed by them over a specific interval. This region can be computed by taking the difference between the two functions at any point within this interval and integrating this difference over the range.
When you have two curves, say, a parabola and a straight line, the area between them might be the shaded part, highlighting the space between the curves rather than the entire area under each one.
The technique systems from the definite integral, where you evaluate the area under the curve from one point to another. To find the area between two curves, you take the upper function minus the lower function throughout the interval of interest.
Definite Integral
The definite integral is a fundamental concept in calculus used to calculate areas under curves across a specific interval. Unlike indefinite integrals, which return general forms or families of functions, definite integrals have limits of integration, providing a specific numerical result.
For example, when looking for the area between two curves like in the provided problem, the definite integral helps by giving the exact area enclosed. You integrate the function formed by the difference of the upper and lower curve over the specified limits.
  • It allows you to determine a fixed area on a graph.
  • The limits of integration signify the interval along the x-axis you're interested in.
Thus, understanding the definite integral helps us solve real-world problems involving area, displacement, and accumulation.
Intersection of Curves
The intersection of curves is vital in determining the limits of integration when finding the area between them. Finding where curves intersect tells you where to start and stop integrating.
To find these intersection points, you set the two function equations equal to each other and solve for the variable. In the case of linear and quadratic functions, like the ones in the original exercise, it often involves setting up a simple algebraic equation and solving for the variable that provides the points where the curves meet.
These solutions will become the boundaries of the definite integral, ensuring you measure the area accurately between the relevant points on each curve.
Power Rule in Integration
The power rule in integration is a straightforward technique essential for finding antiderivatives, especially in polynomial functions. It states that to integrate a term of the form \(x^n\), you add one to the exponent and divide by the new exponent, then add the constant of integration for indefinite integrals.

For example:
  • Integrate \(x^n\) to get \(\frac{x^{n+1}}{n+1} + C\).
  • Apply this to individual terms within a polynomial function.
The constant \(C\) is omitted in definite integrals since the limits of integration ultimately cancel it out.
This rule simplifies integration processes and makes calculations, like finding the area between curves involving polynomial expressions, more straightforward. By understanding and applying the power rule, you can efficiently evaluate integrals and solve calculus problems accurately.

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