(i) The Maclaurin's series is:
1 ₊ x ₊ x²/2! . 1 ₋ 3x⁴/4! ₊ ...........
(ii) Integral of limit o to 1∫ e^sinx is 0
Maclaurin's expansion is as follows:
y(x) = y₀ ₊ x y₁(0) ₊ x²/2! y₂(0) ₊ x³/3! y₃(0) ₊ ........
y = e^sinx
y(0) = e^sin0
y(0) = 1
Differentiate y with respect to x
y₁(x) = e^sinx . cos x
y₁(0) = 1 [i.e y₁ = cosx.y]
y₁(0) = 1
Differentiate y₁ with respect to x
y₂(x) = cosc . y₁ ₋ sinx . y
∴ y₂(0) = 1 ₋ 0 = 1
Differentiate y₂ with respect to x
y₃ = cosx . y₂ ₋ (2 sinx ₊ cosx)y₁
y₃(0) = 1 ₋(0 ₊ 1)1
y₃(0) = 0
Differentiate y₃ with respect to x
y₄ = cosx . y₃ sinx . y₃ ₋ (2sinx ₊ cosx)y₂ ₋ (2cos x ₋ sinx) y₁
y₄(0) = 0 ₋ (3 . 0 ₊ 1)1 ₋ (2 ₋ o) 1
y₄(0) = -3
∴ From maclaurin's equation we get:
e^sinx = 1 ₊ x . 1 ₊ x²/2! . 1 ₊ x³/3! (0) ₊ x⁴/4! (₋3) ₊ ....
=1 ₊ x ₊ x²/2! . 1 ₋ 3x⁴/4! ₊ ...........
Hence we get the the series expansion as 1 ₊ x ₊ x²/2! . 1 ₋ 3x⁴/4! ₊ ...........
(ii) limit 0 to 1∫e^sinx
integration formula ∫e^x dx = e^x + c
limit 0 to 1 ∫e^sinx = [e^sinx + c ]limit 0 to 1
= e^sin0 ₋ e^sin1
= 1 ₋ 1
= 0
Hence we get the answer as 0
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