Lösung 2.1:4a

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Zeile 1: Zeile 1:
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{{NAVCONTENT_START}}
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If we draw the curve
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<center> [[Image:2_1_4a-1(2).gif]] </center>
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<math>y=\sin x</math>, we see that the curve lies above the
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{{NAVCONTENT_STOP}}
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<math>x</math>
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{{NAVCONTENT_START}}
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-axis as far as
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<center> [[Image:2_1_4a-2(2).gif]] </center>
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<math>x=\pi </math>
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{{NAVCONTENT_STOP}}
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and then lies under the
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<math>x</math>
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-axis.
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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
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<math>x=0</math>
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and
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<math>x=\pi </math>
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can therefore be written as
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<math>\int\limits_{0}^{\pi }{\sin x}\,dx</math>
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whilst the area of the remaining region under the
 +
<math>x</math>
 +
-axis is equal to
 +
 +
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<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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 +
 +
<math>\begin{align}
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& \int\limits_{0}^{\pi }{\sin x}\,dx-\int\limits_{\pi }^{{5\pi }/{4}\;}{\sin x}\,dx \\
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& =\left[ -\cos x \right]_{0}^{\pi }-\left[ -\cos x \right]_{\pi }^{{5\pi }/{4}\;} \\
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& =\left( -\cos \pi -\left( -\cos 0 \right) \right)-\left( -\cos \frac{5\pi }{4}-\left( -\cos \pi \right) \right) \\
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& =\left( -\left( -1 \right)-\left( -1 \right) \right)-\left( -\left( -\frac{1}{\sqrt{2}} \right)-\left( -\left( -1 \right) \right) \right) \\
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& =1+1-\frac{1}{\sqrt{2}}+1=3-\frac{1}{\sqrt{2}} \\
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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 \frac{5\pi }{4}</math>
 +
is to draw the angles
 +
<math>0</math>,
 +
<math>\pi </math>
 +
and
 +
<math>\frac{5\pi }{4}</math>
 +
on a unit circle and to read off the cosine value as the
 +
<math>x</math>
 +
-coordinate for the corresponding point on the circle.
 +
 +
[[Image:2_1_4_a2.gif|center]]
[[Image:2_1_4_a2.gif|center]]

Version vom 10:02, 18. Okt. 2008

If we draw the curve \displaystyle y=\sin x, we see that the curve lies above the \displaystyle x -axis as far as \displaystyle x=\pi and then lies under the \displaystyle 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 \displaystyle 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 \\ & =\left[ -\cos x \right]_{0}^{\pi }-\left[ -\cos x \right]_{\pi }^{{5\pi }/{4}\;} \\ & =\left( -\cos \pi -\left( -\cos 0 \right) \right)-\left( -\cos \frac{5\pi }{4}-\left( -\cos \pi \right) \right) \\ & =\left( -\left( -1 \right)-\left( -1 \right) \right)-\left( -\left( -\frac{1}{\sqrt{2}} \right)-\left( -\left( -1 \right) \right) \right) \\ & =1+1-\frac{1}{\sqrt{2}}+1=3-\frac{1}{\sqrt{2}} \\ \end{align}


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


Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:4a“