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$$\int^{x=\frac{\pi}{4}}_{x=0} \int^{y= \cos{x}}_{y=\sin{x}} dydx$$

I got the answer $\sqrt{2} - 1 $ but my tutor got $8$? I assumed that I am starting with integrating $1 dydx$, that is how I got my answer.

Is my answer wrong?

jimjim
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Lils
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  • Your answer seems correct. – Andrés Morales Jun 13 '21 at 05:08
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    show how to get that answer, what is important is not just a result at the end but the way to get it, a wrong answer with a correct way is way better than a correct answer done the wrong way. – jimjim Jun 13 '21 at 05:14
  • Like I said I started with 1dydx then pretty simple integrated f(x,y) = y using cosx-sinx then so on, but I had no idea how my tutor got 8! – Lils Jun 13 '21 at 05:42
  • Your tutor is wrong. The answer is definitely $\sqrt{2}-1$. – Davis Parks Jun 13 '21 at 05:29

1 Answers1

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So we have: $$ I = \int_{x = 0}^{x = \frac{\pi}{4}} \!\!\!\int_{y = \sin{x}}^{y = \cos{x}} \!\!dydx = \int_{x = 0}^{x = \frac{\pi}{4}}\!\! \big(\cos{x} - \sin{x}\big) dx $$

The area bounded between cosine and sine between $0$ and $\frac{\pi}{4}$.

Hence:

$$ I =\big( \sin{x} + \cos{x} \big)_{x = 0}^{x=\frac{\pi}{4}} = \left(\sin{\frac{\pi}{4}} + \cos{\frac{\pi}{4}}\right) - \big(\sin{0} + \cos{0}) \big) = \frac{\sqrt2}2+\frac{\sqrt{2}}{2}- 1 = \sqrt{2} - 1 $$

Your answer is correct!

Angelo
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Andrés Morales
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