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To determine whether the integral [tex]\(\int_2^{\infty} \frac{3}{x^2 + 3x} \, dx\)[/tex] is convergent or divergent, and evaluate it if it's convergent, follow these steps:
1. Rewrite the Integrand:
Rewrite the integrand to simplify it:
[tex]\[ \frac{3}{x^2 + 3x} = \frac{3}{x(x + 3)} \][/tex]
2. Use Partial Fraction Decomposition:
Decompose the rational function into partial fractions to simplify the integration process:
[tex]\[ \frac{3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3} \][/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], set up the equation:
[tex]\[ 3 = A(x + 3) + Bx \][/tex]
Equate coefficients for like terms:
[tex]\[ 3 = A(x + 3) + Bx \implies 3 = Ax + 3A + Bx \implies 3 = (A + B)x + 3A \][/tex]
From this, we get two equations:
[tex]\[ A + B = 0 \][/tex]
[tex]\[ 3A = 3 \][/tex]
Solving these, we find:
[tex]\[ A = 1 \quad \text{and} \quad B = -1 \][/tex]
Thus, the integrand can be written as:
[tex]\[ \frac{3}{x(x + 3)} = \frac{1}{x} - \frac{1}{x + 3} \][/tex]
3. Set Up the Integral:
Substitute the partial fractions back into the integral:
[tex]\[ \int_2^{\infty} \frac{3}{x^2 + 3x} \, dx = \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx \][/tex]
4. Integrate Term by Term:
Integrate each term separately:
[tex]\[ \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx = \int_2^{\infty} \frac{1}{x} \, dx - \int_2^{\infty} \frac{1}{x + 3} \, dx \][/tex]
Integrate each term:
[tex]\[ \int \frac{1}{x} \, dx = \ln|x| \quad \text{and} \quad \int \frac{1}{x+3} \, dx = \ln|x + 3| \][/tex]
Therefore:
[tex]\[ \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx = \left[ \ln|x| - \ln|x + 3| \right]_2^{\infty} \][/tex]
5. Evaluate the Definite Integral:
Evaluate the limits from [tex]\(2\)[/tex] to [tex]\(\infty\)[/tex]:
[tex]\[ \left[ \ln|x| - \ln|x + 3| \right]_2^{\infty} = \lim_{t \to \infty} \left( \ln|t| - \ln|t + 3| \right) - \left( \ln|2| - \ln|5| \right) \][/tex]
As [tex]\(t \to \infty\)[/tex]:
[tex]\[ \ln|t| - \ln|t + 3| \to 0 \quad \text{because} \quad \ln|t(t+3)^{-1}| \approx \ln(1) = 0 \][/tex]
Therefore:
[tex]\[ \left( \ln|2| - \ln|5| \right) = -\ln\left(\frac{5}{2}\right) \][/tex]
Hence, the integral converges, and its value is:
[tex]\[ -(\ln 2 - \ln 5) = -\ln 2 + \ln 5 = \ln\left(\frac{5}{2}\right) \][/tex]
So, the improper integral [tex]\(\int_2^{\infty} \frac{3}{x^2 + 3x} \, dx\)[/tex] is convergent, and its value is [tex]\(-\ln (2) + \ln (5)\)[/tex].
1. Rewrite the Integrand:
Rewrite the integrand to simplify it:
[tex]\[ \frac{3}{x^2 + 3x} = \frac{3}{x(x + 3)} \][/tex]
2. Use Partial Fraction Decomposition:
Decompose the rational function into partial fractions to simplify the integration process:
[tex]\[ \frac{3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3} \][/tex]
To find [tex]\(A\)[/tex] and [tex]\(B\)[/tex], set up the equation:
[tex]\[ 3 = A(x + 3) + Bx \][/tex]
Equate coefficients for like terms:
[tex]\[ 3 = A(x + 3) + Bx \implies 3 = Ax + 3A + Bx \implies 3 = (A + B)x + 3A \][/tex]
From this, we get two equations:
[tex]\[ A + B = 0 \][/tex]
[tex]\[ 3A = 3 \][/tex]
Solving these, we find:
[tex]\[ A = 1 \quad \text{and} \quad B = -1 \][/tex]
Thus, the integrand can be written as:
[tex]\[ \frac{3}{x(x + 3)} = \frac{1}{x} - \frac{1}{x + 3} \][/tex]
3. Set Up the Integral:
Substitute the partial fractions back into the integral:
[tex]\[ \int_2^{\infty} \frac{3}{x^2 + 3x} \, dx = \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx \][/tex]
4. Integrate Term by Term:
Integrate each term separately:
[tex]\[ \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx = \int_2^{\infty} \frac{1}{x} \, dx - \int_2^{\infty} \frac{1}{x + 3} \, dx \][/tex]
Integrate each term:
[tex]\[ \int \frac{1}{x} \, dx = \ln|x| \quad \text{and} \quad \int \frac{1}{x+3} \, dx = \ln|x + 3| \][/tex]
Therefore:
[tex]\[ \int_2^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx = \left[ \ln|x| - \ln|x + 3| \right]_2^{\infty} \][/tex]
5. Evaluate the Definite Integral:
Evaluate the limits from [tex]\(2\)[/tex] to [tex]\(\infty\)[/tex]:
[tex]\[ \left[ \ln|x| - \ln|x + 3| \right]_2^{\infty} = \lim_{t \to \infty} \left( \ln|t| - \ln|t + 3| \right) - \left( \ln|2| - \ln|5| \right) \][/tex]
As [tex]\(t \to \infty\)[/tex]:
[tex]\[ \ln|t| - \ln|t + 3| \to 0 \quad \text{because} \quad \ln|t(t+3)^{-1}| \approx \ln(1) = 0 \][/tex]
Therefore:
[tex]\[ \left( \ln|2| - \ln|5| \right) = -\ln\left(\frac{5}{2}\right) \][/tex]
Hence, the integral converges, and its value is:
[tex]\[ -(\ln 2 - \ln 5) = -\ln 2 + \ln 5 = \ln\left(\frac{5}{2}\right) \][/tex]
So, the improper integral [tex]\(\int_2^{\infty} \frac{3}{x^2 + 3x} \, dx\)[/tex] is convergent, and its value is [tex]\(-\ln (2) + \ln (5)\)[/tex].
Answer:\ln \frac{5}{2}
Step-by-step explanation:
1. Simplify the integrand:
\int_{2}^{\infty} \frac{3}{x^2 + 3x} \, dx = \int_{2}^{\infty} \frac{3}{x(x + 3)} \, dx
2. Partial Fraction Decomposition:
Decompose \frac{3}{x(x + 3)} into partial fractions:
\frac{3}{x(x + 3)} = \frac{A}{x} + \frac{B}{x + 3}
Solving for A and B:
3 = A(x + 3) + Bx
Setting x = 0:
\[
3 = A(0 + 3) \implies A = 1
\]
Setting x = -3:
\[
3 = B(-3) \implies B = -1
\]
Therefore:
\frac{3}{x(x + 3)} = \frac{1}{x} - \frac{1}{x + 3}
3. Rewrite the integral:
\int_{2}^{\infty} \frac{3}{x(x + 3)} \, dx = \int_{2}^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx
4. Integrate term by term:
\int_{2}^{\infty} \left( \frac{1}{x} - \frac{1}{x + 3} \right) \, dx = \int_{2}^{\infty} \frac{1}{x} \, dx - \int_{2}^{\infty} \frac{1}{x + 3} \, dx
Find the antiderivative of each term:
\int \frac{1}{x} \, dx = \ln|x| + C
\int \frac{1}{x + 3} \, dx = \ln|x + 3| + C
5. Evaluate the definite integrals:
\left[ \ln|x| \right]{2}^{\infty} - \left[ \ln|x + 3| \right]{2}^{\infty}
Consider the limit as x \to \infty:
\lim_{b \to \infty} \left( \int_{2}^{b} \frac{1}{x} \, dx - \int_{2}^{b} \frac{1}{x + 3} \, dx \right)
= \lim_{b \to \infty} \left( \left[ \ln|x| \right]{2}^{b} - \left[ \ln|x + 3| \right]{2}^{b} \right)
= \lim_{b \to \infty} \left( \ln b - \ln 2 \right) - \lim_{b \to \infty} \left( \ln (b + 3) - \ln 5 \right)
6. Simplify the expression:
= \lim_{b \to \infty} \left( \ln b - \ln (b + 3) \right) + (\ln 5 - \ln 2)
= \lim_{b \to \infty} \left( \ln \frac{b}{b + 3} \right) + \ln \frac{5}{2}
As b \to \infty, \frac{b}{b + 3} \to 1, and \ln 1 = 0:
= 0 + \ln \frac{5}{2}
= \ln \frac{5}{2}
Therefore, the integral is convergent, and its value is:
\ln \frac{5}{2}
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