(a) Write down the first 4 non-zero terms of the Maclaurin series of
f(x) = sin( 0.1 x^2 )
(b) Use the results of the previous part to estimate
f(x) = ∫ 0 to 1.2 (sin( 0.1x^2 ) dx)
Please show working please
f(x) = sin( 0.1 x^2 )
(b) Use the results of the previous part to estimate
f(x) = ∫ 0 to 1.2 (sin( 0.1x^2 ) dx)
Please show working please
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a) Using the series for sine,
sin(0.1x^2)
= (0.1x^2) - (0.1x^2)^3/3! + (0.1x^2)^5/5! - (0.1x^2)^7/7! + ...
= 0.1x^2 - (0.1)^3 x^6/3! + (0.1)^5 x^10/5! - (0.1)^7 x^14/7! + ...
Integrate each term from 0 to 1.2:
∫(x = 0 to 1.2) sin(0.1x^2) dx
= 0.1x^3/3 - (0.1)^3 x^7/(3! * 7) + (0.1)^5 x^11/(5! * 11) - (0.1)^7 x^15/(7! * 15) + ... {for x = 0 to 1.2}
= 0.1(1.2)^3/3 - (0.1)^3 (1.2)^7/(3! * 7) + (0.1)^5 (1.2)^11/(5! * 11) - (0.1)^7 (1.2)^15/(7! * 15) + ...
Using these four terms, the integral approximately equals 0.0575.
I hope this helps!
sin(0.1x^2)
= (0.1x^2) - (0.1x^2)^3/3! + (0.1x^2)^5/5! - (0.1x^2)^7/7! + ...
= 0.1x^2 - (0.1)^3 x^6/3! + (0.1)^5 x^10/5! - (0.1)^7 x^14/7! + ...
Integrate each term from 0 to 1.2:
∫(x = 0 to 1.2) sin(0.1x^2) dx
= 0.1x^3/3 - (0.1)^3 x^7/(3! * 7) + (0.1)^5 x^11/(5! * 11) - (0.1)^7 x^15/(7! * 15) + ... {for x = 0 to 1.2}
= 0.1(1.2)^3/3 - (0.1)^3 (1.2)^7/(3! * 7) + (0.1)^5 (1.2)^11/(5! * 11) - (0.1)^7 (1.2)^15/(7! * 15) + ...
Using these four terms, the integral approximately equals 0.0575.
I hope this helps!