Which of the following integrals is improper because the interval of integration is infinite?

• A. ${\int }_0^1{x}^{2}dx$
• B. ${\int }_1^2\sqrt{x}dx$
• C. ${\int }_1^{\infty }\frac{1}{x^2}dx$
• D. ${\int }_{-1}^1\frac{1}{x}dx$

Answer: (C)

Verified step by step guidance

1. Step 1: Understand the definition of an improper integral. An integral is considered improper if it involves an infinite interval of integration or if the integrand becomes unbounded within the interval.
2. Step 2: Analyze each integral provided in the problem. For example, int_0^1 x^2 dx has a finite interval [0, 1] and the integrand x^2 is well-behaved (bounded) within this interval, so it is not improper.
3. Step 3: Examine int_1^2 sqrt{x} dx. The interval [1, 2] is finite, and the integrand sqrt{x} is also bounded within this interval, so this integral is not improper.
4. Step 4: Consider int_1^{infty} frac{1}{x^2} dx. The interval of integration is [1, ∞), which is infinite. This makes the integral improper due to the infinite upper limit of integration.
5. Step 5: Review int_{-1}^1 frac{1}{x} dx. The interval [-1, 1] is finite, but the integrand frac{1}{x} becomes unbounded at x = 0 (division by zero). This makes the integral improper due to the unbounded behavior of the integrand within the interval.