Answer:
a
Since the integral has an infinite discontinuity, it is a Type 2 improper integral
b
Since the integral has an infinite interval of integration, it is a Type 1 improper integral
c
Since the integral has an infinite interval of integration, it is a Type 1 improper integral
d
Since the integral has an infinite discontinuity, it is a Type 2 improper integral
Step-by-step explanation:
Considering a
[tex]\int\limits^4_3 {\frac{x}{x- 3} } \, dx[/tex]
Looking at this we that at x = 3 this integral will be infinitely discontinuous
Considering b
[tex]\int\limits^{\infty}_0 {\frac{1}{1 + x^3} } \, dx[/tex]
Looking at this integral we see that the interval is between [tex]0 \ and \ \infty[/tex] which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral
Considering c
[tex]\int\limits^{\infty}_{- \infty} {x^2 e^{-x^2}} \, dx[/tex]
Looking at this integral we see that the interval is between [tex]-\infty \ and \ \infty[/tex] which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral
Considering d
[tex]\int\limits^{\frac{\pi}{4} }_0 {cot (x)} \, dx[/tex]
Looking at the integral we see that at x = 0 cot (0) will be infinity hence the integral has an infinite discontinuity , so it is a Type 2 improper integral
