[6 6n power series representation of 1-t6 25) 6n + 1 G2+1 Now J 1< dt = C + 6n + 1 Submit Skip (You cannot come back) Stcp

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Evaluate the indefinite integral as power series_ What is the radius of convergence R?
1 – [
Step
Using
power series representation of
Step
Since 1 – [6
6n
power series representation of 1-t6 {25)
6n + 1
G2+1
Now
J 1< dt = C +
6n + 1
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Stcp

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01:10

Evaluate the indefinite integral as a power series. What is the radius of convergence R?

Step 1:Using a power series representation of 1 – t^10, we have:∫(1 – t^10) dt = ∫(∑(n=0)^∞ (-1)^n * t^(10n)) dt

Step 2:Since 1 – t^10 can be represented as a power series, we can rewrite the integral as:∫(∑(n=0)^∞ (-1)^n * t^(10n)) dt = ∑(n=0)^∞ (∫((-1)^n * t^(10n)) dt)

Step 3:Evaluating the integral, we get:∫((-1)^n * t^(10n)) dt = ((-1)^n * t^(10n+1))/(10n+1)

Now, the integral becomes:∑(n=0)^∞ ((-1)^n * ((t^(10n+1))/(10n+1)))

What is the radius of convergence R?

02:58

Evaluate the indefinite integral as a power series. What is the radius of convergence?

Using the power series representation of t^10:

1 – t^10 = Σ (10n)(-1)^n t^(10n+1)

Now, let’s find the integral:

∫ dt = C + Σ (10n)(-1)^n t^(10n+2)/(10n+1)

The series for the indefinite integral converges for t^10, which means it has a radius of convergence.

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04:59

Evaluate the indefinite integral as a power series:dt 1 – t3C +n =What is the radius of convergence R? R

01:57

Evaluate the indefinite integral as a power series: ∫(x^8 ln(1 + x)) dxf(x) = C + ∑(n+1)^2/(n+1)(n+10)n = 1

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07:45

indefinite integral as a power series

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