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.

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.