Problem 33
Question
It is often possible to change an improper integral into
a proper one by using integration by parts. Consider \(\lim _{c \rightarrow
0^{+}} \int_{c}^{1} \frac{d x}{\sqrt{x}(1+x)} .\) Use integration by parts on
the interval
\([c, 1]\) where \(0
Step-by-Step Solution
Verified Answer
The improper integral becomes: \( 1 + 2 \int_{0}^{1} \frac{\sqrt{x} \, dx}{(1+x)^2} \).
1Step 1: Identify Functions for Integration by Parts
In integration by parts, identify the functions to use. Let \( u = \frac{1}{1+x} \) so that \( du = -\frac{1}{(1+x)^2} \, dx \) and let \( dv = \frac{dx}{\sqrt{x}} \) with \( v = 2\sqrt{x} \), as the derivative of \( x^{1/2} \) is simpler to integrate.
2Step 2: Apply Integration by Parts Formula
Use the integration by parts formula \( \int u \, dv = uv - \int v \, du \). Here, compute \( uv = \frac{2\sqrt{x}}{1+x} \bigg|_c^1 \) and \( \int v \, du = \int_{c}^{1} \frac{2\sqrt{x}}{(1+x)^2} \, dx \).
3Step 3: Evaluate Boundary Terms
Substitute the boundaries into \( uv \): \( \frac{2\sqrt{x}}{1+x} \bigg|_c^1 = \frac{2\sqrt{1}}{1+1} - \frac{2\sqrt{c}}{c+1} = 1 - \frac{2\sqrt{c}}{c+1} \).
4Step 4: Substitute into Equation
Insert the results from steps 2 and 3 into the integration by parts rearrangement: \( \int_{c}^{1} \frac{dx}{\sqrt{x}(1+x)} = 1 - \frac{2\sqrt{c}}{c+1} + 2 \int_{c}^{1} \frac{\sqrt{x} \, dx}{(1+x)^2} \).
5Step 5: Take the Limit as \(c \rightarrow 0\)
Observe that as \( c \rightarrow 0 \), \( \frac{2\sqrt{c}}{c+1} \rightarrow 0 \), making the problematic boundary term disappear. Therefore, the original improper integral turns into a proper one: \( 1 + 2 \int_{0}^{1} \frac{\sqrt{x} \, dx}{(1+x)^2} \).
Key Concepts
Integration by PartsLimit ProcessDefinite Integrals
Integration by Parts
Integration by parts is a technique used to integrate products of functions. It is derived from the product rule for differentiation. The general formula is:
In our problem, we apply integration by parts by choosing \( u = \frac{1}{1+x} \) and \( dv = \frac{dx}{\sqrt{x}} \).
We need to differentiate \( u \) to find \( du \), which gives us \( du = -\frac{1}{(1+x)^2} \, dx \).
Meanwhile, integrating \( dv \) gives us \( v = 2\sqrt{x} \).
With \( u \) and \( dv \) identified, the next step involves calculating \( uv \) and \( \int v \, du \).
For this particular example, solving this leads to terms that simplify the original integral significantly. In essence, integration by parts bridges the complex relationship of functions, allowing us to manage integrals that are otherwise tough to handle.
- \( \int u \, dv = uv - \int v \, du \)
In our problem, we apply integration by parts by choosing \( u = \frac{1}{1+x} \) and \( dv = \frac{dx}{\sqrt{x}} \).
We need to differentiate \( u \) to find \( du \), which gives us \( du = -\frac{1}{(1+x)^2} \, dx \).
Meanwhile, integrating \( dv \) gives us \( v = 2\sqrt{x} \).
With \( u \) and \( dv \) identified, the next step involves calculating \( uv \) and \( \int v \, du \).
For this particular example, solving this leads to terms that simplify the original integral significantly. In essence, integration by parts bridges the complex relationship of functions, allowing us to manage integrals that are otherwise tough to handle.
Limit Process
The limit process is essential in dealing with improper integrals. These integrals typically have infinite or undefined bounds.
By applying limits, we aim to transform such integrals into definite ones with finite bounds.
In this exercise, our focus is to switch from the limit form \( \lim_{c \to 0^{+}} \int_{c}^{1} \frac{dx}{\sqrt{x}(1+x)} \) to a proper integral.
As \( c \) approaches zero in the limit \( c \to 0^{+} \), any terms dependent on \( c \) become insignificant as \( c \rightarrow 0 \).
This concept becomes evident in Step 5 of the solution, where the term \( \frac{2\sqrt{c}}{c+1} \) approaches zero.
This term initially represented the boundary condition for \( c \), but through the limit process, we observe that it nullifies.
Understanding this allows us to effectively reduce improper integrals to definite forms, making them more practical to evaluate.
By applying limits, we aim to transform such integrals into definite ones with finite bounds.
In this exercise, our focus is to switch from the limit form \( \lim_{c \to 0^{+}} \int_{c}^{1} \frac{dx}{\sqrt{x}(1+x)} \) to a proper integral.
As \( c \) approaches zero in the limit \( c \to 0^{+} \), any terms dependent on \( c \) become insignificant as \( c \rightarrow 0 \).
This concept becomes evident in Step 5 of the solution, where the term \( \frac{2\sqrt{c}}{c+1} \) approaches zero.
This term initially represented the boundary condition for \( c \), but through the limit process, we observe that it nullifies.
Understanding this allows us to effectively reduce improper integrals to definite forms, making them more practical to evaluate.
Definite Integrals
Definite integrals are integrals with specified upper and lower limits, often giving a number as a result.
They represent the area under a curve between two specific bounds on the graph of a function.
Unlike improper integrals, definite integrals do not involve infinity or undefined limits.
In our problem, after using integration by parts and taking the limit, we transform the original integral into a definite form.
The new integral \( 2 \int_{0}^{1} \frac{\sqrt{x} \, dx}{(1+x)^2} \) is a definite integral since it is limited to the bounds from 0 to 1.
The advantage of attaining a definite integral is that it is straightforward to evaluate, yielding exact numerical results.
Ultimately, mastering definite integrals enhances one's ability to resolve areas and bounds in calculus efficiently.
They represent the area under a curve between two specific bounds on the graph of a function.
Unlike improper integrals, definite integrals do not involve infinity or undefined limits.
In our problem, after using integration by parts and taking the limit, we transform the original integral into a definite form.
The new integral \( 2 \int_{0}^{1} \frac{\sqrt{x} \, dx}{(1+x)^2} \) is a definite integral since it is limited to the bounds from 0 to 1.
The advantage of attaining a definite integral is that it is straightforward to evaluate, yielding exact numerical results.
Ultimately, mastering definite integrals enhances one's ability to resolve areas and bounds in calculus efficiently.
Other exercises in this chapter
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