Question

# Explain why each of the following integrals is improper. (a) \int_6^7 \frac{x}{x-6}dx -Since the integral has an infinite interval of integration, it

Applications of integrals
Explain why each of the following integrals is improper.
(a) $$\int_6^7 \frac{x}{x-6}dx$$
-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
-The integral is a proper integral.
(b)$$\int_0^{\infty} \frac{1}{1+x^3}dx$$
Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
The integral is a proper integral.
(c) $$\int_{-\infty}^{\infty}x^2 e^{-x^2}dx$$
-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
-The integral is a proper integral.
d)$$\int_0^{\frac{\pi}{4}} \cot x dx$$
-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
-The integral is a proper integral.

2021-06-13
a)
Consider the integral $$\int_6^7 \frac{x}{x-6}dx$$
It is improper because it is discontinuous at x=6
Since the integral has an infinite discontinuity , it is a type 2 improper integral.
b) Let us consider the integral $$\int_0^{\infty} \frac{dx}{1+x^3}$$
Since the integral has an infinite interval of integration. That is $$(0,\infty)$$
It is Type 1 improper integral.
c)
Consider the integral $$\int_{-\infty}^{\infty} x^2 e^{-x^2}dx$$
Since the integral has an infinite interval of integration. That is $$(-\infty,\infty)$$
It is Type 1 improper integral.
d)
Consider the interval $$\int_0^{\frac{\pi}{4}} \cot x dx$$
It is improper because it is discontinuous at x=0
Since the integral has an infinite discontinuity , it is a type 2 improper integral.