# Calculus II: In this problem, we explore whether for any continuous functions f(x)?

(See picture below for question). For a) I got that the function goes towards infinity and for part b) I said it diverges. Are these answers correct? I'm not sure how to answer c and d.

### 3 Answers

- Anonymous1 month agoFavorite Answer
3. (a)

inf.

I=S (x^3)dx/(x^4+1)=>

0

inf.

I=(1/4)S(4x^3)dx/(x^4+1)=>

0

inf.

I=ln(x^4+1)->infinity.

0

(b)

inf.

I=S (t^3)dx/(t^4+1)=>

-inf.

I=limit [ln(t^4+1)-ln((-t)^4+1)]/4=0 is convergent.

t->inf.

(c) I=0.

(d) true.

- PinkgreenLv 71 month ago
3. (a)

...inf.

I=S (x^3)dx/(x^4+1)=>

...0

..........inf.

I=(1/4)S(4x^3)dx/(x^4+1)=>

...........0

.......inf.

I=ln(x^4+1)->infinity.

........0

(b)

...inf.

I=S (t^3)dx/(t^4+1)=>

..-inf.

I=limit [ln(t^4+1)-ln((-t)^4+1)]/4=0 is convergent.

...t->inf.

(c) I=0.

(d) true.

- ...Show all comments
You may do it in the way you suggested.

- Login to reply the answers

- 1 month ago
the integral in part a doesn't exist since the function heads toward infinity.

correct on part b, the function is divergent so the integral is. When you find the integral with two infinite bounds you have to split it in two, negatve infinity to some vale plus the integral from that value to positive infinity. in this case each of these ntegrals does not exist

part c actually has an answer you wind up with the limit of the difference of two identical functions, so it goes to 0, and the limit of 0 at anywhere is 0. let me know if you don't get that. one part that might mix things up is remembering (-x)^2 = x^2

d is concluding that the two pieces are not equal, so that is not true. infinity is not a number, but more a concept. so while it makes sense that the function to the left of 0 cancels out the function to the right of 0, we have no idea how "fast" those inifnities are growing. A number tells us "we stop here" so we don't need to worry about how fast it grows. i really hope this makes sense. If not let me know.

- Login to reply the answers

Trisha: It looks like you were the person who first solved

this problem. An excellent copier.