1. Express the integral x^2arctan(x^3)dx as a power series and then estimate the same integral from 0 to 1 with an error less than 0.004.
2. How many terms of the Maclaurin series for f(x)=xln(1+x) do you need to use in order to estimate ln(1.5)^(1/2) to within 0.0006.
Please i need this asap even if u can do one question ty
2. How many terms of the Maclaurin series for f(x)=xln(1+x) do you need to use in order to estimate ln(1.5)^(1/2) to within 0.0006.
Please i need this asap even if u can do one question ty
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1) Start with the geometric series 1/(1 - x) = Σ(n = 0 to ∞) x^n.
Let x = -t^2:
1/(1 - (-t^2)) = Σ(n = 0 to ∞) (-t^2)^n
==> 1/(1 + t^2) = Σ(n = 0 to ∞) (-1)^n t^(2n).
Integrate both sides (term by term) from 0 to t:
arctan t = Σ(n = 0 to ∞) (-1)^n t^(2n+1)/(2n+1).
Let t = x^3:
arctan(x^3) = Σ(n = 0 to ∞) (-1)^n x^(6n+3)/(2n+1).
Multiply both sides by x^2:
x^2 arctan(x^3) = Σ(n = 0 to ∞) (-1)^n x^(6n+5)/(2n+1).
Integrate both sides:
∫ x^2 arctan(x^3) dx = [Σ(n = 0 to ∞) (-1)^n x^(6n+6)/((6n+6)(2n+1))] + C.
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Integrating from 0 to 1 yields
∫(x = 0 to 1) x^2 arctan(x^3) dx = Σ(n = 0 to ∞) (-1)^n / ((6n+6)(2n+1)).
Since this is an alternating series, the error after n terms is not bigger than the following (n+1)th term in absolute value. That is, we need
|(-1)^(n+1) / ((6n+12)(2n+3))| < 0.004
==> (6n+12)(2n+3) > 1/0.004 = 250
==> n > 2.8 (by quadratic formula or graphing)
So, we need at least 3 terms for the desired accuracy.
==> ∫(x = 0 to 1) x^2 arctan(x^3) dx
≈ Σ(n = 0 to 3) (-1)^n / ((6n+6)(2n+1)) ≈ 0.144.
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2) Start with the geometric series 1/(1 - x) = Σ(n = 0 to ∞) x^n.
Let x = -t:
1/(1 + t) = Σn = 0 to ∞) (-1)^n t^n.
Integrate both sides (term by term) from 0 to x:
ln(1 + x) = Σ(n = 0 to ∞) (-1)^n x^(n+1)/(n+1).
Multiply both sides by x:
x ln(1 + x) = Σ(n = 0 to ∞) (-1)^n x^(n+2)/(n+1).
Let x = 0.5:
(1/2) ln(1.5) = ln(√(1.5)) = Σ(n = 0 to ∞) (-1)^n (0.5)^(n+2)/(n+1).
Since this is an alternating series, the error after n terms is bounded by
|(-1)^(n+1) (0.5)^(n+3)/(n+2)| < 0.0006
==> (0.5)^(n+3)/(n+2) < 0.0006
==> 2^(n+3) * (n+2) > 1/0.0006
==> n > 4.9.
So, we need to use at least 5 terms.
==> ln(√(1.5)) ≈ Σ(n = 0 to 5) (-1)^n (0.5)^(n+2)/(n+1) ≈ 0.2023.
I hope this helps!
Let x = -t^2:
1/(1 - (-t^2)) = Σ(n = 0 to ∞) (-t^2)^n
==> 1/(1 + t^2) = Σ(n = 0 to ∞) (-1)^n t^(2n).
Integrate both sides (term by term) from 0 to t:
arctan t = Σ(n = 0 to ∞) (-1)^n t^(2n+1)/(2n+1).
Let t = x^3:
arctan(x^3) = Σ(n = 0 to ∞) (-1)^n x^(6n+3)/(2n+1).
Multiply both sides by x^2:
x^2 arctan(x^3) = Σ(n = 0 to ∞) (-1)^n x^(6n+5)/(2n+1).
Integrate both sides:
∫ x^2 arctan(x^3) dx = [Σ(n = 0 to ∞) (-1)^n x^(6n+6)/((6n+6)(2n+1))] + C.
------------
Integrating from 0 to 1 yields
∫(x = 0 to 1) x^2 arctan(x^3) dx = Σ(n = 0 to ∞) (-1)^n / ((6n+6)(2n+1)).
Since this is an alternating series, the error after n terms is not bigger than the following (n+1)th term in absolute value. That is, we need
|(-1)^(n+1) / ((6n+12)(2n+3))| < 0.004
==> (6n+12)(2n+3) > 1/0.004 = 250
==> n > 2.8 (by quadratic formula or graphing)
So, we need at least 3 terms for the desired accuracy.
==> ∫(x = 0 to 1) x^2 arctan(x^3) dx
≈ Σ(n = 0 to 3) (-1)^n / ((6n+6)(2n+1)) ≈ 0.144.
---------------------------
2) Start with the geometric series 1/(1 - x) = Σ(n = 0 to ∞) x^n.
Let x = -t:
1/(1 + t) = Σn = 0 to ∞) (-1)^n t^n.
Integrate both sides (term by term) from 0 to x:
ln(1 + x) = Σ(n = 0 to ∞) (-1)^n x^(n+1)/(n+1).
Multiply both sides by x:
x ln(1 + x) = Σ(n = 0 to ∞) (-1)^n x^(n+2)/(n+1).
Let x = 0.5:
(1/2) ln(1.5) = ln(√(1.5)) = Σ(n = 0 to ∞) (-1)^n (0.5)^(n+2)/(n+1).
Since this is an alternating series, the error after n terms is bounded by
|(-1)^(n+1) (0.5)^(n+3)/(n+2)| < 0.0006
==> (0.5)^(n+3)/(n+2) < 0.0006
==> 2^(n+3) * (n+2) > 1/0.0006
==> n > 4.9.
So, we need to use at least 5 terms.
==> ln(√(1.5)) ≈ Σ(n = 0 to 5) (-1)^n (0.5)^(n+2)/(n+1) ≈ 0.2023.
I hope this helps!