Which of the following integrals are improper? (Select all that apply.)

• (a) $\int_{2}^{3} \frac{1}{3x - 1} \, dx$
• (b) $\int_{0}^{1} \frac{1}{2x - 1} \, dx$
• (c) $\int_{-\infty}^{\infty} \frac{\sin(x)}{1 + 4x^2} \, dx$
• (d) $\int_{1}^{5} \ln(x - 1) \, dx$

Neetesh Kumar | December 10, 2024

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This is the solution to Math 132
Assignment: 7.8 Question Number 2
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Step-by-step solution:

(a) Analyze the integral

The integral is: $\int_{2}^{3} \frac{1}{3x - 1} \, dx$

• The denominator $3x - 1$ does not equal $0$ for any $x$ in the interval $[2, 3]$.
• There are no infinite discontinuities or infinite limits in the interval.

Conclusion: The integral is not improper.

(b) Analyze the integral

The integral is: $\int_{0}^{1} \frac{1}{2x - 1} \, dx$

• The denominator $2x - 1$ becomes $0$ at $x = 0.5$, which lies within the interval of integration $[0, 1]$.
• This creates an infinite discontinuity in the interval.

Conclusion: The integral is a Type 2 improper integral due to the infinite discontinuity at $x = 0.5$.

(c) Analyze the integral

The integral is: $\int_{-\infty}^{\infty} \frac{\sin(x)}{1 + 4x^2} \, dx$

• The limits of integration are $-\infty$ and $\infty$, which makes the interval of integration infinite.

Conclusion: The integral is a Type 1 improper integral due to the infinite interval of integration.

(d) Analyze the integral

The integral is: $\int_{1}^{5} \ln(x - 1) \, dx$

• The argument of the logarithmic function, $x - 1$, becomes $0$ at $x = 1$, which is the lower limit of integration.
• This creates an infinite discontinuity at $x = 1$.

Conclusion: The integral is a Type 2 improper integral due to the infinite discontinuity at $x = 1$.

Final Answer:

The improper integrals are:

(b): $\boxed{\text{Type 2 improper integral.}}$

(c): $\boxed{\text{Type 1 improper integral.}}$

(d): $\boxed{\text{Type 2 improper integral.}}$

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