Area between two curves, trigonometric functions (HSC Question) (1 Viewer)

BlueGas · Apr 30, 2015

I need help 1996 HSC question 4 c) (this question isn't actually from 1996 so I'm not sure why the teacher wrote 1996)

By the way, to see the picture on a proper angle, drag it to a new tab and you can zoom to see the question on a proper angle.

InteGrand · Apr 30, 2015

BlueGas said:
I need help 1996 HSC question 4 c) (this question isn't actually from 1996 so I'm not sure why the teacher wrote 1996)

By the way, to see the picture on a proper angle, drag it to a new tab and you can zoom to see the question on a proper angle.

According to the drawn diagram, the area from $x=0$ to $\frac{\pi}{6}$ is given by $A_1= \int _0 ^{\frac{\pi}{6}}\sin 2x \text{ d}x$ . Also, the area from $x=\frac{\pi}{6}$ to $\frac{\pi}{2}$ is given by $A_2 = \int _{\frac{\pi}{6}} ^{\frac{\pi}{2}}\cos x \text{ d}x$ . So the total area in question is given by $A_\text{total}=A_1 + A_2=\int _0 ^{\frac{\pi}{6}}\sin 2x \text{ d}x +\int _{\frac{\pi}{6}} ^{\frac{\pi}{2}}\cos x \text{ d}x$ .

BlueGas · Apr 30, 2015

InteGrand said:
According to the drawn diagram, the area from $x=0$ to $\frac{\pi}{6}$ is given by $A_1= \int _0 ^{\frac{\pi}{6}}\sin 2x \text{ d}x$ . Also, the area from $x=\frac{\pi}{6}$ to $\frac{\pi}{2}$ is given by $A_2 = \int _{\frac{\pi}{6}} ^{\frac{\pi}{2}}\cos x \text{ d}x$ . So the total area in question is given by $A_\text{total}=A_1 + A_2=\int _0 ^{\frac{\pi}{6}}\sin 2x \text{ d}x +\int _{\frac{\pi}{6}} ^{\frac{\pi}{2}}\cos x \text{ d}x$ .

I understand it perfectly now, thank you so much!

By the way, are you a teacher/ student at highschool/ student at uni?

Area between two curves, trigonometric functions (HSC Question) (1 Viewer)

BlueGas

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InteGrand

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BlueGas

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