Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# 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. Relevance
• Anonymous
1 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.

• Pinkgreen
Lv 7
1 month agoReport

Trisha: It looks like you were the person who first solved
this problem. An excellent copier.

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• 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.

• Pinkgreen
Lv 7
1 month agoReport

You may do it in the way you suggested.

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• 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.

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