Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 2.1:4a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.1:4a moved to Solution 2.1:4a: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
If we draw the curve
-
<center> [[Image:2_1_4a-1(2).gif]] </center>
+
<math>y=\sin x</math>, we see that the curve lies above the
-
{{NAVCONTENT_STOP}}
+
<math>x</math>
-
{{NAVCONTENT_START}}
+
-axis as far as
-
<center> [[Image:2_1_4a-2(2).gif]] </center>
+
<math>x=\pi </math>
-
{{NAVCONTENT_STOP}}
+
and then lies under the
 +
<math>x</math>
 +
-axis.
 +
 
[[Image:2_1_4_a1.gif|center]]
[[Image:2_1_4_a1.gif|center]]
 +
 +
The area of the region between
 +
<math>x=0</math>
 +
and
 +
<math>x=\pi </math>
 +
can therefore be written as
 +
 +
 +
<math>\int\limits_{0}^{\pi }{\sin x}\,dx</math>
 +
 +
whilst the area of the remaining region under the
 +
<math>x</math>
 +
-axis is equal to
 +
 +
 +
<math>-\int\limits_{\pi }^{{5\pi }/{4}\;}{\sin x}\,dx</math>
 +
 +
 +
(note the minus sign in front of the integral).
 +
 +
The total area becomes
 +
 +
 +
<math>\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}</math>
 +
 +
 +
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 y=sinx, we see that the curve lies above the x -axis as far as x= and then lies under the x -axis.


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


0sinxdx 

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


−54sinxdx 


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

The total area becomes


0sinxdx−54sinxdx=−cosx0−−cosx54=−cos−−cos0−−cos45−−cos=−−1−−1−−−12−−−1=1+1−12+1=3−12


NOTE: a simple way to obtain the values of cos0, cos and cos45 is to draw the angles 0, and 45 on a unit circle and to read off the cosine value as the x -coordinate for the corresponding point on the circle.


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