Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Use the Trapezoidal Rule with \(n=4\) to approximate the definite integral. $$ \int_{-1}^{1} \frac{1}{x^{2}+1} d x $$

Short Answer

Expert verified
The approximation of the definite integral using the Trapezoidal Rule with 4 trapezoids is 1.75.

Step by step solution

01

Calculate the width of the trapezoids

The width \( h \) of each trapezoid in the approximation is the range of the integral divided by the number of trapezoids. In general, this can be formulated as \( h = \frac{b - a}{n} \), where [a, b] is the interval and n is the number of trapezoids. In this case: \( h = \frac{1 -(-1)}{4} = 0.5 \)
02

Find the x-values of the trapezoids

We need to calculate the x-values where the function is evaluated. Start at the beginning of the interval and proceed upwards, adding the width of the trapezoids until the end of the interval is reached: the x-values are -1, -0.5, 0, 0.5, 1.
03

Calculate the function's values at the x

Substitute x-values into the function \( f(x) = \frac{1}{x^{2} + 1} \). We find the values to be approximately: \( f(-1) = 0.5, f(-0.5) \approx 0.8, f(0) = 1, f(0.5) \approx 0.8, f(1) = 0.5 \).
04

Apply trapezoidal rule to approximate the integral

Now, apply the formula: \( \int_{a}^{b} f(x) dx \approx \frac{h}{2} [f(x_{0}) + 2f(x_{1}) + 2f(x_{2}) + . . . + 2f(x_{n-1}) + f(x_{n})] \). Substituting in the pre calculated values gives an approximation of \( \frac{0.5}{2} [0.5 + 2*0.8 + 2*1 + 2*0.8 + 0.5] = 1.75 \).

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 Vaia!

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

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.

Sign-up for free