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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]
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