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Chapter 5: Problem 293

In the following exercises, use a change of variables to evaluate the definite integral. $$ \int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} d x $$

Short Answer

Expert verified
The integral evaluates to \( \sqrt{2} - 1 \).

Step by step solution

01

Identify the substitution

To solve the integral \( \int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} \, dx \), we will use a substitution method. Notice that the denominator involves \( 1+x^2 \), suggesting we can use \( u = 1 + x^2 \). This simplifies the square root function and its derivative can be easily matched with parts of the integrand.
02

Calculate the derivative and differential

Differentiate \( u = 1 + x^2 \) to get \( \frac{du}{dx} = 2x \). Solving for \( dx \), we obtain \( dx = \frac{du}{2x} \). This substitute will help in changing the variable of integration from \( x \) to \( u \).
03

Change the limits of integration

The original limits are from \( x = 0 \) to \( x = 1 \). Use the substitution to find the new limits. When \( x = 0 \), \( u = 1 + 0^2 = 1 \). When \( x = 1 \), \( u = 1 + 1^2 = 2 \). So, the limits change to \( u \) from 1 to 2.
04

Substitute and simplify the integral

Substitute \( u = 1 + x^2 \) and \( dx = \frac{du}{2x} \) into the integral: \[ \int_{1}^{2} \frac{x}{\sqrt{u}} \cdot \frac{du}{2x}. \] The \( x \) terms cancel out, simplifying to \[ \frac{1}{2} \int_{1}^{2} \frac{1}{\sqrt{u}} \, du. \]
05

Evaluate the integral

Now, solve \( \frac{1}{2} \int_{1}^{2} u^{-1/2} \, du \). The antiderivative of \( u^{-1/2} \) is \( 2u^{1/2} \). So \( \frac{1}{2} \cdot 2u^{1/2} = u^{1/2} \). Evaluate this from 1 to 2: \[ \left[ u^{1/2} \right]_{1}^{2} = 2^{1/2} - 1^{1/2}. \]
06

Simplify the final answer

Compute the expression \( 2^{1/2} - 1^{1/2} = \sqrt{2} - 1 \). This is the value of the integral after evaluating it from 1 to 2.

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Key Concepts

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

definite integrals
Definite integrals represent the area under a curve between two specific points, on a graph. In the integral notation \( \int_a^b f(x) \, dx \), \( a \) and \( b \) are the limits of integration. These limits define the interval on the x-axis over which the function \( f(x) \) is integrated.

The main idea is to calculate the exact area bounded by the function and the x-axis between these two points. Because definite integrals provide an area, they have a specific numerical value, unlike indefinite integrals that have a constant of integration.

A few important points about definite integrals:
  • The value of a definite integral could be positive, zero, or negative, depending on the function's position relative to the x-axis.
  • They do not include the constant of integration \( C \) since they result in a specific numeric value rather than a general form.
  • In practice, they often require special integration techniques, such as substitution or integration by parts, to solve.
substitution method
The substitution method is a common technique used in integration to simplify the evaluation of integrals. This technique involves changing variables to transform a complex integral into a simpler one.

The purpose is to find a suitable substitution that simplifies the integrand, making it easier to integrate. Here鈥檚 how it is typically done:
  • Identify a part of the integrand that can be rewritten using a new variable (often denoted \( u \)), simplifying its differentiation.
  • Differentiate this new variable to determine \( du \), aiding in transforming the differential part of the integral \( dx \) in terms of \( du \).
  • Rewrite the integrand in terms of \( u \) and \( du \) and then solve the simpler integral.
In our original exercise, we chose the substitution \( u = 1 + x^2 \) since the derivative \( 2x \) matches a part of the integrand. The substitution method simplifies solving the integral by removing complex expressions or inconvenient forms.
change of variables
The change of variables is a fundamental concept in calculus, particularly useful in integration. It allows us to modify the variable of integration, changing the appearance of the integrand to something more manageable.

When applying a change of variables:
  • The original limits of integration are recalculated to fit the new variable. This involves substituting the limits through the chosen function.
  • The integrand is rewritten in terms of the new variable, potentially simplifying the expression significantly.
  • Solving the rewritten integral yields results that can be directly evaluated based on the new limits.
In this exercise, after choosing \( u = 1 + x^2 \), we recalculate the limits: originally from \( x = 0 \) to \( x = 1 \), which translates to \( u = 1 \) to \( u = 2 \). This new form often makes evaluating the integral more straightforward.
antiderivative
An antiderivative of a function is a function whose derivative is the original function. Finding antiderivatives is a crucial part of solving integrals, particularly indefinite integrals.

To solve an integral by finding its antiderivative, follow these steps:
  • Understand the form of the original function, considering basic antiderivatives.
  • Apply necessary integration techniques, possibly finding a substitution or using known identities.
  • For definite integrals, calculate the antiderivative first, then compute its value at the various limits.
In our exercise, after making the variable change, we found the antiderivative of \( u^{-1/2} \), which is \( 2u^{1/2} \). This step transforms the integral into a direct evaluation task over the new limits, from \( u = 1 \) to \( u = 2 \). The difference of these evaluations gives the solution.

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Most popular questions from this chapter

In the following exercises, use the identity \(\int_{-A}^{A} f(x) d x=\int_{-A}^{0} f(x) d x+\int_{0}^{A} f(x) d x\) to compute the integrals. \(\int_{1}^{3}(2-x) d x(\text { Hint: Look at the graph of } f .)\)

Show that the average value of \(\sin ^{2} t\) over \([0,2 \pi]\) is equal to 1\(/ 2\) Without further calculation, determine whether the average value of \(\sin ^{2} t\) over \([0, \pi]\) is also equal to 1\(/\)2.

In the following exercises, compute the average value using the left Riemann sums \(L_{N}\) for \(N=1,10,100\) . How does the accuracy compare with the given exact value? \([\mathrm{T}] \quad y=x e^{x^{2}}\) over the interval \([0,2] ;\) the exact solution is \(\frac{1}{4}\left(e^{4}-1\right)\)

In the following exercises, use the comparison theorem. Show that \(\int_{0}^{\pi / 2} \sin t d t \geq \frac{\pi}{4} .\) (Hint: sin \(t \geq \frac{2 t}{\pi}\) over \(\left[0, \frac{\pi}{2}\right])\)

Evaluate the Riemann sums \(L_{4}\) and \(R_{4}\) for the following functions over the specified interval. Compare your answer with the exact answer, when possible, or use a calculator to determine the answer. $$y=x^{2} \sin x \text { over }[0, \pi]$$
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