Question

Which of following integrals are improper ? Why? (a) \(\int_{1}^{2} \frac{1}{2 x-1} \mathrm{dx}\) (b) \(\int_{0}^{1} \frac{1}{2 x-1} d x\) (c) \(\int_{-\infty}^{\infty} \frac{\sin x}{1+x^{2}} d x\) (d) \(\int_{1}^{2} \ln (x-1) d x\)

Short answer

a) \(\int_{1}^{2} \frac{1}{2x-1} \mathrm{dx}\) b) \(\int_{0}^{1} \frac{1}{2x-1} \mathrm{dx}\) c) \(\int_{-\infty}^{\infty} \frac{\sin x}{1+x^2} \mathrm{dx}\) d) \(\int_{1}^{2} \ln(x-1) \mathrm{dx}\) Answer: Integrals b, c, and d are improper. Integral b is improper due to the discontinuity within the interval at \(x = \frac{1}{2}\), integral c is improper because of infinite integration limits, and integral d is improper due to the discontinuity and unboundedness within the interval at \(x=1\).

Step-by-step solution

Let's check the conditions for the first integral: \(\int_{1}^{2} \frac{1}{2x-1} \mathrm{dx}.\) Both integration limits (1 and 2) are finite, the function is continuous and there is no point where it is unbounded within the range (1, 2). Therefore, integral (a) is a proper integral. #Step 2: Analyze the second integral (b)#

Let's check the conditions for the second integral: \(\int_{0}^{1} \frac{1}{2x-1} \mathrm{dx}.\) Both integration limits (0 and 1) are finite, but there is a discontinuity at \(x = \frac{1}{2}\) since the denominator becomes 0. Therefore, integral (b) is an improper integral due to the discontinuity within the interval. #Step 3: Analyze the third integral (c)#

Let's check the conditions for the third integral: \(\int_{-\infty}^{\infty} \frac{\sin x}{1+x^2} \mathrm{dx}.\) The integration limits are infinite (-∞ and ∞) which makes this integral improper. The function inside the integral is continuous throughout its domain, but the infinite limits automatically make this integral improper. #Step 4: Analyze the fourth integral (d)#

Let's check the conditions for the fourth integral: \(\int_{1}^{2} \ln(x-1) \mathrm{dx}.\) Both integration limits (1 and 2) are finite. However, there is a discontinuity (and unboundedness) at \(x=1\) as \(\ln(x-1)\) becomes indeterminate (\(-\infty\)) at that point. Thus, integral (d) is an improper integral due to discontinuity and unboundedness within the interval.