How do you do this question?

0.001591

Step-by-step explanation:

The power series for arctan(x) is:

arctan(x) = ∑ (-1)ⁿ x²ⁿ⁺¹ / (2n + 1)

Substituting 5x:

arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺¹ / (2n + 1)

Multiply both sides by x:

x arctan(5x) = ∑ (-1)ⁿ x (5x)²ⁿ⁺¹ / (2n + 1)

Simplify:

x arctan(5x) = ∑ (-1)ⁿ (5x) (5x)²ⁿ⁺¹ / (10n + 5)

x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺² / (10n + 5)

Multiply top and bottom by 5:

x arctan(5x) = ∑ (-1)ⁿ 5 (5x)²ⁿ⁺² / (50n + 25)

Integrate:

∫ x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))

Evaluate between x = 0.1 and x = 0:

∫₀⁰¹ x arctan(5x) = [∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))] |₀⁰¹

∫₀⁰¹ x arctan(5x) = ∑ (-1)ⁿ (0.5)²ⁿ⁺³ / ((50n + 25) (2n + 3))

This is an alternating series.  We can approximate it with Alternating Series Estimation.

bₙ₊₁ ≥ ε

(0.5)²⁽ⁿ⁺¹⁾⁺³ / ((50(n+1) + 25) (2(n+1) + 3)) ≥ 0.000001

(0.5)²ⁿ⁺⁵ / ((50n + 75) (2n + 5)) ≥ 0.000001

n ≥ 3

So the approximation is the sum of the terms from n=0 to n=3.

(-1)⁰ (0.5)²⁽⁰⁾⁺³ / ((50(0) + 25) (2(0) + 3))

+ (-1)¹ (0.5)²⁽¹⁾⁺³ / ((50(1) + 25) (2(1) + 3))

+ (-1)² (0.5)²⁽²⁾⁺³ / ((50(2) + 25) (2(2) + 3))

+ (-1)³ (0.5)²⁽³⁾⁺³ / ((50(3) + 25) (2(3) + 3))

= 0.0016667 − 0.0000833 + 0.0000089 − 0.0000012

= 0.001591

b is the final answer! : )

hope this .

r3t40