First, construct the sixth degree Taylor polynomial P6(x) for function f(x)=sin⁡(x2) about x0=0 The use ∫01P6(x)dx to approximate the integral ∫01sin⁡(x2)dx. Use 4-digit rounding arithmetic in all calculations. What is the approximate value?

Question

First, construct the sixth degree Taylor polynomial P6(x) for function f(x)=sin⁡(x2) about x0=0 The use ∫01P6(x)dx to approximate the integral  ∫01sin⁡(x2)dx. Use 4-digit rounding arithmetic in all calculations. What is the approximate value?

Clara Reese · Accepted Answer

Taylor series f(x)=sin⁡(x2) is =f(x0)+(x−x0)1!f′(x0)+(x−x0)22!f(x0)±−−− Note that x0=0 So, the polynomial isf(0)+xf′(0)+x22f&quot;(0)+x36f′(0)+x424f′(0)+x5120f′(0)+x6720⋅f6(0)+−− (up to sixth degree) Now f(x)=sin⁡(x2)f′(x)=cos⁡(x2)⋅2xfx)=2[cos⁡x2+x⋅sin⁡(x2)(−2x)]=2[cos⁡x2−2x2sin⁡x2]f(x)=2(−sin⁡(x2)⋅2x)−4[x2cos⁡x2⋅2x+sin⁡x2⋅2x]=−4xsin⁡(x2)−8x3cos⁡x2−8xsin⁡x2=−12xsin⁡x2−8x3cos⁡x2f4(x)=ddx(f(x))=−12(sin⁡x2+x⋅cos⁡x2 ⋅2x)−8(cos⁡x2⋅3x2−x3⋅sin⁡x2⋅2x)

Jeffrey Jordon · Accepted Answer

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First, construct the sixth degree Taylor polynomial P_6(x) for function f(x) = sin(x^2) about x_0 = 0 The use int_{0}^{1} P_6 (x) dx to approximate the integral int_{0}^{1} sin(x^2)dx. Use 4-digit rounding arithmetic in all calculations. What is the approximate value?

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