Help with Cambridge Questions (1 Viewer)

dragon658 · Mar 26, 2017

Can someone please go through question 7a and 7b page 190 of Cambridge Year 12 Second Edition (orange and blue book)

I've got the exam tomorrow!

Drongoski · Mar 26, 2017

Provide the questions - not every potential helper has the book.

dan964 · Mar 26, 2017

questions for OP request

1729 · Mar 26, 2017

dragon658 said:
Can someone please go through question 7a and 7b page 190 of Cambridge Year 12 Second Edition (orange and blue book)

I've got the exam tomorrow!

$$\noindent \textbf{(7a)} Now the intersection point is at $x = \frac{\pi}{4}$ and so the area $A = \int_0^{\frac{\pi}{4}}\sin{x}\mathrm{d}x + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos{x}\mathrm{d}x.$ But by symmetry, these two areas are equal and thus we can just write $A = 2 \times \int_0^{\frac{\pi}{4}}\sin{x}\mathrm{d}x = 2\left[-\cos{x}\right]_0^{\frac{\pi}{4}} = 2\left[-\cos{\frac{\pi}{4}} + \cos{0}\right] = 2\left[1-\frac{1}{\sqrt{2}}\right] = 2\left[1 - \frac{\sqrt{2}}{2}\right] = \left(2 - \sqrt{2}\right)$ square units.$

$$\noindent \textbf{(7b)} Here, we divide the area into two, so we have $A = \int_{-1}^0\left(x+1\right)\mathrm{d}x + \int_0^{\frac{\pi}{2}} \cos{x}\mathrm{d}x$. But $\int_{-1}^0\left(x+1\right)\mathrm{d}x$ simply represents the area of a triangle with base and height of 1. And so $A = \frac{1}{2}bh + \int_0^{\frac{\pi}{2}} \cos{x}\mathrm{d}x = \frac{1}{2}(1)(1) + \left[\sin{x}\right]_0^{\frac{\pi}{2}} = \frac{1}{2} + \left[\sin{\frac{\pi}{2}} - \sin{0}\right] = \frac{1}{2} + 1 = 1\frac{1}{2}$ square units.$

Help with Cambridge Questions (1 Viewer)

dragon658

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Drongoski

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dan964

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1729

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