/*! 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 92 Evaluate the following integrals... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 92

Evaluate the following integrals. $$\int_{25}^{225} \frac{d x}{\sqrt{x^{2}+25 x}}(\text { Hint: } \sqrt{x^{2}+25 x}=\sqrt{x} \sqrt{x+25} .)$$

Short Answer

Expert verified
Question: Evaluate the following integral using the given hint: $$\int_{25}^{225} \frac{dx}{\sqrt{x^2+25x}},$$ where $$\sqrt{x^2+25x} = \sqrt{x}\sqrt{x+25}.$$ Answer: The given integral evaluates to 40.

Step by step solution

01

Apply the given hint

Rewrite the integral using the provided hint. Substitute \(\sqrt{x^2+25x}\) with \(\sqrt{x}\sqrt{x+25}\) in the denominator. $$\int_{25}^{225} \frac{dx}{\sqrt{x}\sqrt{x+25}}$$
02

Simplify the integral

Perform a variable change and use substitution to simplify the integral. Let the substitution be \(u=x+25\), then the differential is \(du=dx\). Now the integral becomes: $$\int_{50}^{250} \frac{du}{\sqrt{u(u-25)}}$$ The limits of integration have been changed to account for the change in variables.
03

Perform the integration

To solve the integral, separate the integrand and rewrite it as a product: $$\int_{50}^{250} \frac{du}{\sqrt{u(u-25)}} = \int_{50}^{250} \frac{du}{\sqrt{u}} \cdot \frac{1}{\sqrt{u-25}}$$ Now, observe that: $$\int \frac{du}{\sqrt{u}} \cdot \frac{1}{\sqrt{u-25}} = \int \frac{d(u-25)}{\sqrt{u-25}} \cdot \frac{1}{\sqrt{u}}$$ Thus we can change the variable back to \(x\) and find the antiderivative as the sum of the two separated integrals: $$\int_{25}^{225} \frac{dx}{\sqrt{x}\sqrt{x+25}} = \int_{25}^{225}\left(\frac{dx}{\sqrt x} + \frac{dx}{\sqrt{x+25}}\right)$$
04

Evaluate the separated integrals

Integrate each term separately: $$\int_{25}^{225} \frac{dx}{\sqrt{x}} = \int_{25}^{225} x^{-\frac{1}{2}} dx = \left[2x^{\frac{1}{2}}\right]_{25}^{225}$$ $$\int_{25}^{225} \frac{dx}{\sqrt{x+25}} = \int_{25}^{225} (x+25)^{-\frac{1}{2}} dx = \left[2(x+25)^{\frac{1}{2}}\right]_{25}^{225}$$ Now we evaluate each term at the limits of integration: $$\left[2x^{\frac{1}{2}}\right]_{25}^{225} = 2(15)-2(5) = 20$$ $$\left[2(x+25)^{\frac{1}{2}}\right]_{25}^{225} = 2(15+5)-2(5+5) = 20$$
05

Calculate the final result

Finally, add the results of the two separated integrals: $$\int_{25}^{225} \frac{dx}{\sqrt{x}\sqrt{x+25}} = 20+20 = 40$$ The final value of the given integral is 40.

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.

Definite Integrals
Definite integrals serve as the foundation for understanding several key aspects of calculus. They let us find the accumulation of quantities, such as areas under curves, over a specified interval \(a, b\). In this context, our problem involves evaluating a definite integral:
  • The integral \(\int_{25}^{225} \frac{dx}{\sqrt{x} \sqrt{x+25}}\) means we find the total accumulation from 25 to 225 of the expression inside the integral.
  • When solving a definite integral, the result depends on both the function and the specific interval of integration, from 25 to 225 in our exercise.
  • The definite integral outputs a specific numerical value once evaluated, signifying the total change or area pertaining to the problem's limits.
An important detail about definite integrals is that they have boundaries, which gives a fixed real number, unlike indefinite integrals which result in families of functions.
Substitution Method
The substitution method is a powerful technique to simplify integrals, especially when they seem too complex to handle directly. In our exercise, we used substitution by setting \(u = x + 25\). Let's explore how and why this method works step-by-step:
  • Substitution replaces a difficult part of the integral with a simpler variable, transforming the problem into an easier one.
  • For instance, with the substitution \(u = x + 25\), we altered the variable to make the operations straightforward, converting the integrand into a more manageable form:
Switching from x to u simplifies the integral, making it easier to solve. We also adjusted the limits of integration, from \(x = 25, 225\) to \(u = 50, 250\), maintaining the integrity of the definite integral.
Antiderivative
An antiderivative of a function is another function whose derivative gives the original function. In this context, it is crucial as it allows us to evaluate integrals:
  • Finding the antiderivatives \(\int x^{-\frac{1}{2}} dx\) and \(\int (x+25)^{-\frac{1}{2}} dx\) provides the foundation to solve the definite integral.
  • The antiderivative process is reversed differentiation, where we deduce the original function before differentiation.
  • For example, the antiderivative of \(x^{-\frac{1}{2}}\) is \([2x^{\frac{1}{2}}]\), indicating the function whose derivative reverts to our integrand's form.
Finding antiderivatives allows us to evaluate the expression over the interval and apply the limits to calculate the integral accurately.
Integration Techniques
Integration techniques like substitution and separating integrals are crucial for successfully solving complex integrals. Let's delve into these handy methods:
  • Separation of integrals involves breaking down a complex integral into simpler parts, each solvable independently.
  • Such techniques facilitate tackling sections individually, making the problem more manageable.
  • In this exercise, we split \(\int \frac{dx}{\sqrt{x} \sqrt{x+25}}\) into two simpler integrals involving individual square roots.
These techniques are essential not only for arriving at a solution but also for enhancing your understanding of the integral's behavior and aiding further applications in calculus. Mastering these can greatly boost your ability to approach integrals with confidence and accuracy.

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

At Earth's surface, the acceleration due to gravity is approximately \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) (with local variations). However, the acceleration decreases with distance from the surface according to Newton's law of gravitation. At a distance of \(y\) meters from Earth's surface, the acceleration is given by where \(R=6.4 \times 10^{6} \mathrm{m}\) is the radius of Earth. a. Suppose a projectile is launched upward with an initial velocity of \(v_{0} \mathrm{m} / \mathrm{s} .\) Let \(v(t)\) be its velocity and \(y(t)\) its height (in meters) above the surface \(t\) seconds after the launch. Neglecting forces such as air resistance, explain why \(\frac{d v}{d t}=a(y)\) and \(\frac{d y}{d t}=v(t)\) b. Use the Chain Rule to show that \(\frac{d v}{d t}=\frac{1}{2} \frac{d}{d y}\left(v^{2}\right)\) c. Show that the equation of motion for the projectile is \(\frac{1}{2} \frac{d}{d y}\left(v^{2}\right)=a(y),\) where \(a(y)\) is given previously. d. Integrate both sides of the equation in part (c) with respect to \(y\) using the fact that when \(y=0, v=v_{0} .\) Show that $$ \frac{1}{2}\left(v^{2}-v_{0}^{2}\right)=g R\left(\frac{1}{1+y / R}-1\right) $$ e. When the projectile reaches its maximum height, \(v=0\) Use this fact to determine that the maximum height is \(y_{\max }=\frac{R v_{0}^{2}}{2 g R-v_{0}^{2}}\) f. Graph \(y_{\max }\) as a function of \(v_{0} .\) What is the maximum height when \(v_{0}=500 \mathrm{m} / \mathrm{s}, 1500 \mathrm{m} / \mathrm{s},\) and \(5 \mathrm{km} / \mathrm{s} ?\) g. Show that the value of \(v_{0}\) needed to put the projectile into orbit (called the escape velocity) is \(\sqrt{2 g R}\)

Define the relative growth rate of the function \(f\) over the time interval \(T\) to be the relative change in \(f\) over an interval of length \(T\): $$R_{T}=\frac{f(t+T)-f(t)}{f(t)}.$$ Show that for the exponential function \(y(t)=y_{0} e^{k t},\) the relative growth rate \(R_{T}\) is constant for any \(T ;\) that is, choose any \(T\) and show that \(R_{r}\) is constant for all \(t\).

A hemispherical bowl of radius 8 inches is filled to a depth of \(h\) inches, where \(0 \leq h \leq 8\). Find the volume of water in the bowl as a function of \(h\). (Check the special cases \(h=0 \text { and } h=8 .)\)

The U.S. government reports the rate of inflation (as measured by the Consumer Price Index) both monthly and annually. Suppose that for a particular month, the monthly rate of inflation is reported as \(0.8 \%\). Assuming that this rate remains constant, what is the corresponding annual rate of inflation? Is the annual rate 12 times the monthly rate? Explain.

Use the following argument to show that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\). a. Make a sketch of the function \(f(x)=1 / x\) on the interval \([1,2] .\) Explain why the area of the region bounded by \(y=f(x)\) and the \(x\) -axis on [1,2] is \(\ln 2\) b. Construct a rectangle over the interval [1,2] with height \(\frac{1}{2}\) Explain why \(\ln 2>\frac{1}{2}\). c. Show that \(\ln 2^{n}>n / 2\) and \(\ln 2^{-n}<-n / 2\). d. Conclude that \(\lim _{x \rightarrow \infty} \ln x=\infty\) and \(\lim _{x \rightarrow 0^{+}} \ln x=-\infty\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Calculus

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.