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 table of integrals to find the exact area of the region bounded by the graphs of the equations. Then use a graphing utility to graph the region and approximate the area. $$ y=\frac{x}{\sqrt{x+1}}, y=0, x=8 $$

Short Answer

Expert verified
The exact area under the curve \(y = \frac{x}{\sqrt{x+1}}\) from x = 0 to x = 8 is 12 square units.

Step by step solution

01

Expression of the Problem as a Definite Integral

Since the area under the curve of a non-negative, continuous function from a to b can be calculated as the definite integral of the function from a to b, we can express the problem as follows: \( \int_0^8 \frac{x}{\sqrt{x+1}} dx \) .
02

Computation of the Definite Integral

The computation of the integral can be simplified by using the substitution method. We set \(u = x + 1\). Then, \(du = dx\), \(x = u -1\), and we adjust the limits of the integral to match the new variable. The integral then becomes \( \int_1^9 \frac{u-1}{\sqrt{u}} du \). This integral can be computed as: \( \int_1^9 (u^{1/2} - u^{-1/2}) du \).
03

Solving the Integral

Now we can split the integral and solve step by step: The integral of \(u^{1/2}\) with respect to u is \(\frac{2}{3}u^{3/2}\) and the integral of \(u^{-1/2}\) with respect to u is \(2u^{1/2}\). The resulting calculation brings us to \[\frac{2}{3}u^{3/2} - 2u^{1/2} |_1^9\] .
04

Substituting Upper and Lower Boundaries

We can now substitute the upper and lower boundaries. The exact area under the curve then will be: \[\frac{2}{3}* 9^{3/2} - 2*9^{1/2} - (\frac{2}{3}*1^{3/2} - 2*1^{1/2}) \]
05

Computing the Numerical Result

Finally we can come up with the exact number for the area under the curve: \[18 - 6 - (\frac{2}{3} - 2) = 12\]

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

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{3} \frac{1}{x^{2} \sqrt{x^{2}-9}} d x $$

Approximate the integral using (a) the Trapezoidal Rule and (b) Simpson's Rule for the indicated value of \(n\). (Round your answers to three significant digits.) $$ \int_{0}^{1} \frac{1}{1+x^{2}} d x, n=4 $$

Use a program similar to the Simpson's Rule program on page 906 with \(n=6\) to approximate the indicated normal probability. The standard normal probability density function is \(f(x)=(1 / \sqrt{2 \pi}) e^{-x^{2} / 2}\). If \(x\) is chosen at random from a population with this density, then the probability that \(x\) lies in the interval \([a, b]\) is \(P(a \leq x \leq b)=\int_{a}^{b} f(x) d x\). $$ P(0 \leq x \leq 1) $$

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges, and check your results with the results obtained by using the integration capabilities of a graphing utility. $$ \int_{3}^{4} \frac{1}{\sqrt{x^{2}-9}} d x $$

Use a program similar to the Simpson's Rule program on page 906 with \(n=6\) to approximate the indicated normal probability. The standard normal probability density function is \(f(x)=(1 / \sqrt{2 \pi}) e^{-x^{2} / 2}\). If \(x\) is chosen at random from a population with this density, then the probability that \(x\) lies in the interval \([a, b]\) is \(P(a \leq x \leq b)=\int_{a}^{b} f(x) d x\). $$ P(0 \leq x \leq 2) $$

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