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Estimation of Trapezium Rule



I'm a little stuck on part C. I have worked out part B to be 0.720.

I do remember that Sin^2x = 1 - Cos^2x, so I did 1 - 0.720. I'm not getting the correct answer however.

Any help is appreciated - Thanks
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Zacken
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Original post by CrazyFool229


I'm a little stuck on part C. I have worked out part B to be 0.720.

I do remember that Sin^2x = 1 - Cos^2x, so I did 1 - 0.720. I'm not getting the correct answer however.

Any help is appreciated - Thanks


Yeah, so in the first few parts you saidn ∫0π/3cos⁡2x dx=a\int_0^{\pi/3} \cos^2 x \, \mathrm{d}x = a.

In the last part you need to do ∫0π/31−cos⁡2x dx=x]0π/3−a\int_0^{\pi/3} 1 - \cos^2 x \, \mathrm{d}x = x\big]_0^{\pi/3} - a.

a is your answer to the previous parts and you need to remember that you need to integrate the 1.
Original post by Zacken
Yeah, so in the first few parts you saidn ∫0π/3cos⁡2x dx=a\int_0^{\pi/3} \cos^2 x \, \mathrm{d}x = a.

In the last part you need to do ∫0π/31−cos⁡2x dx=x]0π/3−a\int_0^{\pi/3} 1 - \cos^2 x \, \mathrm{d}x = x\big]_0^{\pi/3} - a.

a is your answer to the previous parts and you need to remember that you need to integrate the 1.


Thanks again Zacken :biggrin:

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