Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To determine whether the given improper integral is convergent or divergent, and if it is convergent, to evaluate it, we consider the integral:
[tex]$ \int_2^{\infty} \frac{3}{x^2 + 3x} \, dx $[/tex]
### Step 1: Simplify the Integrand
First, let's simplify the integrand. Notice that the denominator can be factored:
[tex]$ x^2 + 3x = x(x + 3) $[/tex]
Thus, the integrand becomes:
[tex]$ \frac{3}{x^2 + 3x} = \frac{3}{x(x + 3)} $[/tex]
### Step 2: Use Partial Fraction Decomposition
We can use partial fraction decomposition to split the fraction into simpler components. We assume:
[tex]$ \frac{3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3} $[/tex]
Multiplying both sides by [tex]\( x(x + 3) \)[/tex] to clear the denominators:
[tex]$ 3 = A(x + 3) + Bx $[/tex]
Setting up the system of equations by equating the coefficients on both sides:
1. For the constant term: [tex]\( 3A = 3 \)[/tex]
2. For the coefficient of [tex]\( x \)[/tex]: [tex]\( A + B = 0 \)[/tex]
From the first equation, we get:
[tex]$ A = 1 $[/tex]
From the second equation, we can solve for [tex]\( B \)[/tex]:
[tex]$ 1 + B = 0 \implies B = -1 $[/tex]
So, the partial fraction decomposition is:
[tex]$ \frac{3}{x(x + 3)} = \frac{1}{x} - \frac{1}{x + 3} $[/tex]
### Step 3: Integrate Term-by-Term
Now, rewrite the integral using the partial fractions:
[tex]$ \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx $[/tex]
Separate the integral into two simpler integrals:
[tex]$ \int_2^{\infty} \frac{1}{x} \, dx - \int_2^{\infty} \frac{1}{x + 3} \, dx $[/tex]
### Step 4: Evaluate Each Integral
For the first integral:
[tex]$ \int_2^{\infty} \frac{1}{x} \, dx = \left[ \ln|x| \right]_2^{\infty} $[/tex]
As [tex]\( x \to \infty \)[/tex], [tex]\( \ln|x| \)[/tex] also approaches infinity, thus we need to consider the limit:
[tex]$ \lim_{b \to \infty} \left( \ln b - \ln 2 \right) $[/tex]
For the second integral, make a substitution [tex]\( u = x + 3 \)[/tex]:
[tex]$ \int_2^{\infty} \frac{1}{x + 3} \, dx = \int_5^{\infty} \frac{1}{u} \, du = \left[ \ln|u| \right]_5^{\infty} $[/tex]
### Step 5: Calculate the Definite Integrals
Evaluate the limits:
[tex]$ \int_2^{\infty} \frac{1}{x} \, dx = \left( \lim_{b \to \infty} \ln b - \ln 2 \right) = \lim_{b \to \infty} \ln b - \ln 2 $[/tex]
Here, [tex]\( \ln b \)[/tex] approaches infinity as [tex]\( b \to \infty \)[/tex].
For the second part:
[tex]$ \int_2^{\infty} \frac{1}{x+3} \, dx = \left( \lim_{b \to \infty} \ln(x + 3) - \ln(5) \right) = \lim_{b \to \infty} \ln(b + 3) - \ln 5 $[/tex]
Here, [tex]\( \ln(b + 3) \)[/tex] also approaches infinity as [tex]\( b \to \infty \)[/tex].
### Step 6: Combine the Results
Combining, we get:
[tex]\[ \left( \ln b - \ln 2 \right) - \left( \ln b - \ln 5 \right) \][/tex]
This simplifies as:
[tex]\[ ( \ln b - \ln 2 ) - ( \ln b - \ln 5 ) = - \ln 2 + \ln 5 \][/tex]
Therefore:
[tex]\[ \int_2^{\infty} \frac{3}{x(x+3)} \, dx = \ln 5 - \ln 2 \][/tex]
Since [tex]\( \ln 5 - \ln 2 = \log \frac{5}{2} \)[/tex],
Hence, the integral converges and the final value is:
[tex]$ \int_2^{\infty} \frac{3}{x(x + 3)} \, dx = \ln 5 - \ln 2 $[/tex]
[tex]$ \int_2^{\infty} \frac{3}{x^2 + 3x} \, dx $[/tex]
### Step 1: Simplify the Integrand
First, let's simplify the integrand. Notice that the denominator can be factored:
[tex]$ x^2 + 3x = x(x + 3) $[/tex]
Thus, the integrand becomes:
[tex]$ \frac{3}{x^2 + 3x} = \frac{3}{x(x + 3)} $[/tex]
### Step 2: Use Partial Fraction Decomposition
We can use partial fraction decomposition to split the fraction into simpler components. We assume:
[tex]$ \frac{3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3} $[/tex]
Multiplying both sides by [tex]\( x(x + 3) \)[/tex] to clear the denominators:
[tex]$ 3 = A(x + 3) + Bx $[/tex]
Setting up the system of equations by equating the coefficients on both sides:
1. For the constant term: [tex]\( 3A = 3 \)[/tex]
2. For the coefficient of [tex]\( x \)[/tex]: [tex]\( A + B = 0 \)[/tex]
From the first equation, we get:
[tex]$ A = 1 $[/tex]
From the second equation, we can solve for [tex]\( B \)[/tex]:
[tex]$ 1 + B = 0 \implies B = -1 $[/tex]
So, the partial fraction decomposition is:
[tex]$ \frac{3}{x(x + 3)} = \frac{1}{x} - \frac{1}{x + 3} $[/tex]
### Step 3: Integrate Term-by-Term
Now, rewrite the integral using the partial fractions:
[tex]$ \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx $[/tex]
Separate the integral into two simpler integrals:
[tex]$ \int_2^{\infty} \frac{1}{x} \, dx - \int_2^{\infty} \frac{1}{x + 3} \, dx $[/tex]
### Step 4: Evaluate Each Integral
For the first integral:
[tex]$ \int_2^{\infty} \frac{1}{x} \, dx = \left[ \ln|x| \right]_2^{\infty} $[/tex]
As [tex]\( x \to \infty \)[/tex], [tex]\( \ln|x| \)[/tex] also approaches infinity, thus we need to consider the limit:
[tex]$ \lim_{b \to \infty} \left( \ln b - \ln 2 \right) $[/tex]
For the second integral, make a substitution [tex]\( u = x + 3 \)[/tex]:
[tex]$ \int_2^{\infty} \frac{1}{x + 3} \, dx = \int_5^{\infty} \frac{1}{u} \, du = \left[ \ln|u| \right]_5^{\infty} $[/tex]
### Step 5: Calculate the Definite Integrals
Evaluate the limits:
[tex]$ \int_2^{\infty} \frac{1}{x} \, dx = \left( \lim_{b \to \infty} \ln b - \ln 2 \right) = \lim_{b \to \infty} \ln b - \ln 2 $[/tex]
Here, [tex]\( \ln b \)[/tex] approaches infinity as [tex]\( b \to \infty \)[/tex].
For the second part:
[tex]$ \int_2^{\infty} \frac{1}{x+3} \, dx = \left( \lim_{b \to \infty} \ln(x + 3) - \ln(5) \right) = \lim_{b \to \infty} \ln(b + 3) - \ln 5 $[/tex]
Here, [tex]\( \ln(b + 3) \)[/tex] also approaches infinity as [tex]\( b \to \infty \)[/tex].
### Step 6: Combine the Results
Combining, we get:
[tex]\[ \left( \ln b - \ln 2 \right) - \left( \ln b - \ln 5 \right) \][/tex]
This simplifies as:
[tex]\[ ( \ln b - \ln 2 ) - ( \ln b - \ln 5 ) = - \ln 2 + \ln 5 \][/tex]
Therefore:
[tex]\[ \int_2^{\infty} \frac{3}{x(x+3)} \, dx = \ln 5 - \ln 2 \][/tex]
Since [tex]\( \ln 5 - \ln 2 = \log \frac{5}{2} \)[/tex],
Hence, the integral converges and the final value is:
[tex]$ \int_2^{\infty} \frac{3}{x(x + 3)} \, dx = \ln 5 - \ln 2 $[/tex]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
