Solution 2.1:4a

From Förberedande kurs i matematik 2

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If we draw the curve <math>y=\sin x</math>, we see that the curve lies above the ''x''-axis as far as <math>x=\pi </math> and then lies under the ''x''-axis.
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<center> [[Image:2_1_4a-1(2).gif]] </center>
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[[Image:2_1_4_a1.gif|center]]
[[Image:2_1_4_a1.gif|center]]
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The area of the region between <math>x=0</math> and <math>x=\pi</math> can therefore be written as
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{{Displayed math||<math>\int\limits_{0}^{\pi} \sin x\,dx</math>}}
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whilst the area of the remaining region under the ''x''-axis is equal to
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{{Displayed math||<math>-\int\limits_{\pi}^{5\pi/4} \sin x\,dx</math>}}
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(note the minus sign in front of the integral).
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The total area becomes
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{{Displayed math||<math>\begin{align}
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& \int\limits_{0}^{\pi} \sin x\,dx - \int\limits_{\pi}^{5\pi/4} \sin x\,dx\\[5pt]
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&\qquad\quad {}= \Bigl[\ -\cos x\ \Bigr]_0^\pi - \Bigl[\ -\cos x\ \Bigr]_\pi^{5\pi/4}\\[5pt]
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&\qquad\quad {}= \Bigl( -\cos\pi - (-\cos 0)\Bigr) - \Bigl( -\cos\frac{5\pi}{4} - (-\cos\pi) \Bigr)\\[5pt]
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&\qquad\quad {}= \Bigl( -(-1)-(-1) \Bigr) - \Bigl( -\Bigl(-\frac{1}{\sqrt{2}} \Bigr) - \bigl(-(-1)\bigr)\Bigr)\\[5pt]
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&\qquad\quad {}= 1+1-\frac{1}{\sqrt{2}}+1\\[5pt]
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&\qquad\quad {}= 3-\frac{1}{\sqrt{2}}\,\textrm{.}
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\end{align}</math>}}
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Note: A simple way to obtain the values of <math>\cos 0</math>, <math>\cos \pi </math> and <math>\cos (5\pi/4)</math> is to draw the angles <math>0</math>, <math>\pi</math> and <math>5\pi/4</math> on a unit circle and to read off the cosine value as the ''x''-coordinate for the corresponding point on the circle.
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[[Image:2_1_4_a2.gif|center]]
[[Image:2_1_4_a2.gif|center]]

Current revision

If we draw the curve \displaystyle y=\sin x, we see that the curve lies above the x-axis as far as \displaystyle x=\pi and then lies under the x-axis.

The area of the region between \displaystyle x=0 and \displaystyle x=\pi can therefore be written as

\displaystyle \int\limits_{0}^{\pi} \sin x\,dx

whilst the area of the remaining region under the x-axis is equal to

\displaystyle -\int\limits_{\pi}^{5\pi/4} \sin x\,dx

(note the minus sign in front of the integral).

The total area becomes

\displaystyle \begin{align}

& \int\limits_{0}^{\pi} \sin x\,dx - \int\limits_{\pi}^{5\pi/4} \sin x\,dx\\[5pt] &\qquad\quad {}= \Bigl[\ -\cos x\ \Bigr]_0^\pi - \Bigl[\ -\cos x\ \Bigr]_\pi^{5\pi/4}\\[5pt] &\qquad\quad {}= \Bigl( -\cos\pi - (-\cos 0)\Bigr) - \Bigl( -\cos\frac{5\pi}{4} - (-\cos\pi) \Bigr)\\[5pt] &\qquad\quad {}= \Bigl( -(-1)-(-1) \Bigr) - \Bigl( -\Bigl(-\frac{1}{\sqrt{2}} \Bigr) - \bigl(-(-1)\bigr)\Bigr)\\[5pt] &\qquad\quad {}= 1+1-\frac{1}{\sqrt{2}}+1\\[5pt] &\qquad\quad {}= 3-\frac{1}{\sqrt{2}}\,\textrm{.} \end{align}


Note: A simple way to obtain the values of \displaystyle \cos 0, \displaystyle \cos \pi and \displaystyle \cos (5\pi/4) is to draw the angles \displaystyle 0, \displaystyle \pi and \displaystyle 5\pi/4 on a unit circle and to read off the cosine value as the x-coordinate for the corresponding point on the circle.

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