/*! 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 166 In the following exercises, use ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 166

In the following exercises, use a calculator to estimate the area under the curve by computing \(T_{10}\) , the average of the left- and right-endpoint Riemann sums using \(N=10\) rectangles. Then, using the Fundamental Theorem of Calculus, Part 2 , determine the exact area. $$ y=\sqrt{x^{3}} \text { over }[0,6] $$

Short Answer

Expert verified
The estimated area is \( T_{10} \) and the exact area is \( \frac{2}{5} \times 6^{5/2} \).

Step by step solution

01

Divide the Interval

First, we divide the interval [0, 6] into 10 equal subintervals. Since the interval is from 0 to 6, each subinterval will have a width, or step size, of \( \Delta x = \frac{6 - 0}{10} = 0.6 \).
02

Calculate Left-Endpoint Riemann Sum

The left-endpoint Riemann sum uses the function values at the left endpoint of each interval. Therefore, the sum is given by \( L_{10} = \sum_{i=0}^{9} f(x_i) \cdot \Delta x \) where \( x_i = i \times \Delta x \). Calculate: \( L_{10} = \sum_{i=0}^{9} \sqrt{(i \times 0.6)^3} \times 0.6 \).
03

Calculate Right-Endpoint Riemann Sum

The right-endpoint Riemann sum uses the function values at the right endpoint of each interval. Therefore, the sum is \( R_{10} = \sum_{i=1}^{10} f(x_i) \cdot \Delta x \) where \( x_i = i \times \Delta x \). Calculate: \( R_{10} = \sum_{i=1}^{10} \sqrt{(i \times 0.6)^3} \times 0.6 \).
04

Compute the Average of Left and Right Sums

Now compute the average of the left and right Riemann sums: \( T_{10} = \frac{L_{10} + R_{10}}{2} \). This gives us the estimated area under the curve using the trapezoidal rule.
05

Apply Fundamental Theorem of Calculus

The exact area under the curve can be found using integration. The integral is: \( \int_0^6 \sqrt{x^3} \; dx = \int_0^6 x^{3/2} \; dx \). Find the antiderivative: \( F(x) = \frac{2}{5}x^{5/2} \).
06

Evaluate the Definite Integral

Evaluate \([F(x)]_0^6 = F(6) - F(0) = \frac{2}{5} \times 6^{5/2} - \frac{2}{5} \times 0^{5/2} \). Compute this to obtain the exact area.

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.

Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is an essential concept in calculus that connects differentiation and integration. It comprises two main parts:
  • First Part: It tells us that the derivative of the integral of a function over an interval results in the original function itself. Essentially, differentiation and integration are inverse operations.
  • Second Part: This part helps us evaluate definite integrals. It states that if you have an antiderivative of a function, you can compute the definite integral by analyzing the antiderivative's values at the boundary points of the interval.
In our exercise, to find the exact area under the curve of the function \( y = \sqrt{x^3} \) over the interval [0, 6], we used the second part of the theorem. We first identified the antiderivative \( F(x) = \frac{2}{5}x^{5/2} \). Then, we evaluated this from 0 to 6 to find the exact area beneath the curve.
Definite Integral
A definite integral calculates the total accumulation or the area under a curve of a function over a specific interval. The integral sign, along with upper and lower limits of integration, define the area we are interested in. The function inside the integral is called the integrand, and it is continuously calculated over the interval.
For the function \( y = \sqrt{x^3} \), the definite integral \[ \int_0^6 \sqrt{x^3} \; dx \]translates to finding the area under the curve from \( x = 0 \) to \( x = 6 \). This process involves:
  • Calculating the Antiderivative: The antiderivative of \( x^{3/2} \) is \( \frac{2}{5}x^{5/2} \).
  • Evaluating the Antiderivative at the Interval Boundaries: We calculate \( F(6) \) and \( F(0) \) and subtract these values, which results in the absolute area under the curve within [0, 6].
This computation provides a precise value of the definite integral, reflecting the exact area under the curve.
Trapezoidal Rule
The trapezoidal rule is a numerical method used to estimate the area under a curve. It is helpful when finding an exact integral solution might be complex or infeasible. The method divides the area under a curve into several trapezoids rather than rectangles, as in Riemann sums.
Here's how it works:
  • Divide the Interval: Split the total interval into smaller subintervals.
  • Calculate the Heights: Use function values at both ends of each subinterval to determine the trapezoid's height.
  • Sum the Areas: Combine the areas of all trapezoids to estimate the total area under the curve.
In our calculation, the average of the left and right endpoint Riemann sums \[ T_{10} = \frac{L_{10} + R_{10}}{2} \]provides the trapezoidal estimate for the area under the function \( y = \sqrt{x^3} \) over [0, 6]. This estimate comes from calculating the left \( (L_{10}) \) and right \( (R_{10}) \) endpoint sums separately and then averaging them to account for the curve's slope more accurately. This method offers an efficient way to achieve an area approximation, especially when combined with computational tools like a calculator.

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

In the following exercises, use averages of values at the left (L) and right (R) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals. \(\\{(-4,0),(-2,2),(0,0),(1,2),(3,2),(4,0)\\}\) over \([-4,4]\)

Find the antiderivative. $$\int \frac{4 x^{2}}{\sqrt{1-x^{6}}} d x$$

In the following exercises, approximate the average value using Riemann sums \(L_{100}\) and \(R_{100} .\) How does your answer compare with the exact given answer? [T] \(y=e^{x / 2}\) over the interval \([0,1] ;\) the exact solution is 2\((\sqrt{e}-1)\)

Suppose that \(\int_{0}^{4} f(x) d x=5\) and \(\int_{0}^{2} f(x) d x=-3,\) and \(\int_{0}^{4} g(x) d x=-1\) and \(\int_{0}^{2} g(x) d x=2 .\) In the following exercises, compute the integrals. \(\int_{0}^{2}(f(x)-g(x)) d x\)

[T] Compute the left and right Riemann sums \(L_{10}\) and \(R_{10}\) and their average \(\frac{L_{10}+R_{10}}{2}\) for \(f(t)=t^{2}\) over \([0,1] .\) Given that \(\int_{0}^{1} t^{2} d t=0 . \overline{33}, \quad\) to how many decimal places is \(\frac{L_{10}+R_{10}}{2}\) accurate?
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

Decision 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.