/*! 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 28 Use geometry (not Riemann sums) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use geometry (not Riemann sums) to compute the integrals. HINT [See Quick Examples page 982.]

Short Answer

Expert verified
The given function \(f(x)=\frac{x}{2}\) represents a straight line passing through the origin with a slope of \(\frac{1}{2}\). Within the interval \([-2, 2]\), the curve forms a triangle with the x-axis. We can compute the area of the triangle using the formula \(Area =\frac{1}{2} \times Base \times Height\). For the triangle formed, the base is 4 units and the height is 1 unit. Thus, the area of the triangle is 2 units. Therefore, the integral \(\int_{-2}^{2} \frac{x}{2} d x = 2\).

Step by step solution

01

Determine the shape

The given function is \(f(x)=\frac{x}{2}\), which is a straight line passing through the origin with a slope of \(\frac{1}{2}\). Within the interval \([-2, 2]\), the curve forms a triangle with the x-axis.
02

Calculate the area of the shape

Since the shape formed is a triangle, we'll calculate the area using the formula for the area of a triangle: Area \(=\frac{1}{2} \times Base \times Height\) The base of the triangle is the length of the interval, which is \((-2)-2=4\). The height of the triangle is the maximum value of the function within the interval, which is \(f(2)=\frac{2}{2}=1\). Area \(=\frac{1}{2} \times 4 \times 1 =2\)
03

Use the properties of integrals to calculate the final answer

Now that we have found the area of the triangle, we can compute the integral: \(\int_{-2}^{2} \frac{x}{2} d x = Area\) The area we found in Step 2 is 2 units, which is the answer to the integral: \(\int_{-2}^{2} \frac{x}{2} d x = 2\)

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 91Ó°ÊÓ!

Key Concepts

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

Geometric Interpretation of Integrals
When we talk about integrals, it might sound a bit complex, but one of the friendliest ways to understand them is through geometry. Essentially, integrals can be visualized as the area under a curve on a graph. If we have a function, say \( f(x) = \frac{x}{2} \), and we want to find the integral from \( -2 \) to \( 2 \), we're looking at the area beneath the line this function forms and above the x-axis between these two points.
For the problem \( \int_{-2}^{2} \frac{x}{2} \ dx \), the curve is a straight line and can be interpreted as a geometric shape, here it's a triangle. The integral then represents the area of this triangle. By converting problems into shapes, we often get clearer and more intuitive solutions without the need for complex calculations. This geometric perspective is particularly useful when functions create simple shapes like triangles, rectangles, or circles.
Area under a Curve
Finding the area under a curve sounds technical, but it's often simpler than it appears. The area between the curve \( f(x) \) and the x-axis within a certain interval can often be seen as familiar geometrical shapes. By identifying shapes like triangles or rectangles, we can calculate their areas with basic geometry formulas.
In our specific problem with \( f(x) = \frac{x}{2} \), the area under the curve from \(-2\) to \(2\) forms a triangle with the x-axis. This triangle's base runs horizontally from \(-2\) to \(2\), giving it a length of \(4\). The height is determined by evaluating the function at its highest point in the interval, \(x = 2\), which results in a height of \(1\) (since \( f(2) = \frac{2}{2} = 1 \)).
Using the triangle area formula \( \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \), we find the area to be \( \frac{1}{2} \times 4 \times 1 = 2 \). This results in 2 units squared being the area, which directly corresponds to the value of the integral.
Integration Techniques
In integral calculus, there are numerous techniques to solve integrals, with choice largely dependent on the specific function. While Riemann sums and integration by parts are common, geometric interpretation is especially efficient for functions that correspond to simple geometrical shapes.
This method proves advantageous when our function forms a known shape, like a triangle or rectangle, because we can apply basic geometric formulas instead of more complex integration processes. In our exercise with \( \int_{-2}^{2} \frac{x}{2} \ dx \), recognizing the shape under the curve as a triangle allowed us to quickly compute the integral using the triangle area formula, skipping more advanced calculus techniques.
  • Understanding the shape of the function aids in choosing the most efficient solution method.
  • Recognizing when a simple geometric formula can replace complex calculus saves time and reduces errors.
This approach underscores the importance of analyzing the function graphically before diving into algebraic methods, as it might lead directly to solutions that are both clever and straightforward.

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

The normal distribution curve, which models the distributions of data in a wide range of applications, is given by the function $$ p(x)=\frac{1}{\sqrt{2 \pi} \sigma} e^{-(x-\mu)^{2} / 2 \sigma^{2}} $$ where \(\pi=3.14159265 \ldots\) and \(\sigma\) and \(\mu\) are constants called the standard deviation and the mean, respectively. Its graph (when \(\sigma=1\) and \(\mu=2\) ) is shown in the figure. Illustrate its use. In a survey, consumers were asked to rate a new toothpaste on a scale of \(1-10\). The resulting data are modeled by a normal distribution with \(\mu=4.5\) and \(\sigma=1.0\). The percentage of consumers who rated the toothpaste with a score between \(a\) and \(b\) on the test is given by $$ \int_{a}^{b} p(x) d x $$ a. Use a Riemann sum with \(n=10\) to estimate the percentage of customers who rated the toothpaste 5 or above. (Use the range \(4.5\) to \(10.5\).) b. What percentage of customers rated the toothpaste 0 or \(1 ?\) (Use the range \(-0.5\) to \(1.5\).)

The work done in accelerating an object from velocity \(v_{0}\) to velocity \(v_{1}\) is given by $$ W=\int_{v_{0}}^{v_{1}} v \frac{d p}{d v} d v $$ where \(p\) is its momentum, given by \(p=m v(m=\) mass \()\). Assuming that \(m\) is a constant, show that $$ W=\frac{1}{2} m v_{1}^{2}-\frac{1}{2} m v_{0}^{2} $$ The quantity \(\frac{1}{2} m v^{2}\) is referred to as the kinetic energy of the object, so the work required to accelerate an object is given by its change in kinetic energy.

(Compare Exercise 57 in Section 13.3.) The rate of oil production by Pemex, Mexico's national oil company, can be approximated by \(p(t)=-8.03 t^{2}+73 t+1,060\) million barrels per year, $$ (1 \leq t \leq 9) $$ where \(t\) is time in years since the start of \(2000 .{ }^{39}\) Use a definite integral to estimate total production of oil from the start of 2001 to the start of 2009 .

Use the Fundamental Theorem of Calculus to show that if \(m(x)\) is the marginal cost at a production level of \(x\) items, then the cost function \(C(x)\) is given by $$ C(x)=C(0)+\int_{0}^{x} m(t) d t $$ What do we call \(C(0)\) ?

Calculate the total area of the regions described. Do not count area beneath the \(x\) -axis as negative. HINT [See Example 6.] Bounded by the graph of \(y=|3 x-2|\), the \(x\) -axis, and the lines \(x=0\) and \(x=3\)

See all solutions

Recommended explanations on Math Textbooks

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.