Calc 2- Integrals and Sequences

[notgoodatmath777]

Sorry this is long, I'm mostly confused on parts b and c but would appreciate checking correctness on all parts.


a) I=0.
I did a leading order term analysis and saw that the integrand behave like 1/x^2 which converges by a p-test with p=2>1 and converges to 0 because as x approaches infinity, the denominator gets larger and larger making the numerator very small.

b) Integration by parts form is uv - integral of u'v.
u=(1+x^2)^n u' = 2nx(1+x^2)^(n-1) v'=1. v=x
I solved it out but I cannot seem to figure out where the -1/2n came from which is the uv part. I got uv = -x(1+x^2)^n

c) x^2 = A + B(1+x^2) I multiplied both sides by (1+x^2)^(n+1)
But I cannot find values of x for which A or B is equal to 0 to solve the other.

d) I=0 I feel like it's wrong
I plugged n=3 into the original equation to get the integral of 1/(1+x^2)^3. = 1/(1+x^6). and I did the same thing as part 1 to get that it converges to 0 by the p-test with p=6>1 because denominator gets very large so the numerator gets smaller.

 

[Steven G]

How did you do I₀? What formula did you use? I ask because I_n was only defined for n>1. Please be more careful.

b) If you show us your work we can point out your error.

c) Clearly B must be 1 in order to get the x^2 term. So A + B = A + 1 = ?, so A =?

Again, how can we find the error(s) in your work, if there are any, if you fail to show us your work.

If you had followed the forum's posting guidelines you would have received help by now.

 

[Dr.Peterson]

Sorry this is long, I'm mostly confused on parts b and c but would appreciate checking correctness on all parts.

View attachment 22819
a) I=0.
I did a leading order term analysis and saw that the integrand behave like 1/x^2 which converges by a p-test with p=2>1 and converges to 0 because as x approaches infinity, the denominator gets larger and larger making the numerator very small.

I'm not sure what you're doing here. You were asked to find [MATH]I_1[/MATH], not some series and not some number called [MATH]I[/MATH]. It's just an integral, namely [MATH]\int_1^\infty\frac{dx}{1+x^2}[/MATH] Can you do that?

For each part, please show the details of your work, not just a description.

 

[notgoodatmath777]

I mean I sub 1 is equal to 0.

Could you explain why B=1

 

[Dr.Peterson]

I mean I sub 1 is equal to 0.

Could you explain why B=1
View attachment 22875View attachment 22876

But [MATH]I_1[/MATH] is not zero. Carry out the integral!

For part (c), you have [MATH]x^2 = A + B(1+x^2)[/MATH], which must be true for all x. So the polynomials [MATH]x^2[/MATH] and [MATH]Bx^2 + (A+B)[/MATH] must have the same coefficients, which implies that [MATH]B = 1[/MATH] and [MATH]A+B = 0[/MATH].

You evidently have only learned a shortcut, and not this general method.

 

[Steven G]

In part b why did you suddenly change the signs?

It also seems that in b you claim that (a+b)^n = a^n + b^n +2ab. This is only true for n=2.