Loading http://wiki.math.se/jsMath/fonts/msam10/def.js
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 2.1:4e

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.1:4e moved to Solution 2.1:4e: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
The double inequality means that we look for the area of the region which is bounded above in the y-direction by the straight line
-
<center> [[Image:2_1_4e-1(3).gif]] </center>
+
<math>y=x+\text{2}</math>
-
{{NAVCONTENT_STOP}}
+
and from below by the parabola
-
{{NAVCONTENT_START}}
+
<math>y=x^{2}</math>.
-
<center> [[Image:2_1_4e-2(3).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
If we sketch the line and the parabola, the region is given by the region shaded in the figure below.
-
{{NAVCONTENT_START}}
+
 
-
<center> [[Image:2_1_4e-3(3).gif]] </center>
+
-
{{NAVCONTENT_STOP}}
+
[[Image:2_1_4_e.gif|center]]
[[Image:2_1_4_e.gif|center]]
 +
 +
 +
As soon as we have determined the
 +
<math>x</math>
 +
-coordinates of the points of intersection,
 +
<math>x=a</math>
 +
and
 +
<math>x=b</math>, between the line and the parabola, we can calculate the area as the integral of the difference between the curves'
 +
<math>y</math>
 +
-values:
 +
 +
Area=
 +
<math>\int\limits_{a}^{b}{\left( x+2-x^{2} \right)}\,dx</math>
 +
 +
 +
The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations:
 +
 +
 +
<math>\left\{ \begin{matrix}
 +
y=x+\text{2} \\
 +
y=x^{2} \\
 +
\end{matrix} \right.</math>
 +
 +
 +
By eliminating
 +
<math>y</math>, we obtain an equation for
 +
<math>x</math>,
 +
 +
 +
<math>x^{2}=x+2</math>
 +
 +
 +
If we move all x-terms to the left-hand side,
 +
 +
 +
<math>x^{2}-x=2</math>
 +
 +
 +
and complete the square, we obtain
 +
 +
 +
<math>\begin{align}
 +
& \left( x-\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=2 \\
 +
& \left( x-\frac{1}{2} \right)^{2}=\frac{9}{4} \\
 +
\end{align}</math>
 +
 +
Taking the root then gives that
 +
<math>x=\frac{1}{2}\pm \frac{3}{2}</math>. In other words,
 +
<math>x=-\text{1}</math>
 +
and
 +
<math>x=\text{2}</math>.
 +
 +
The area of the region is now given by
 +
 +
 +
<math>\begin{align}
 +
& \text{Area}=\int\limits_{-1}^{2}{\left( x+2-x^{2} \right)}\,dx=\left[ \frac{x^{2}}{2}+2x-\frac{x^{3}}{3} \right]_{-1}^{2} \\
 +
& =\frac{2^{2}}{2}+2\centerdot 2-\frac{2^{3}}{3}-\left( \frac{\left( -1 \right)^{2}}{2}+2\centerdot \left( -1 \right)-\frac{\left( -1 \right)^{3}}{3} \right) \\
 +
& =2+4-\frac{8}{3}-\frac{1}{2}+2-\frac{1}{3} \\
 +
& =\frac{9}{2} \\
 +
\end{align}</math>

Version vom 13:05, 18. Okt. 2008

The double inequality means that we look for the area of the region which is bounded above in the y-direction by the straight line y=x+2 and from below by the parabola y=x2.

If we sketch the line and the parabola, the region is given by the region shaded in the figure below.



As soon as we have determined the x -coordinates of the points of intersection, x=a and x=b, between the line and the parabola, we can calculate the area as the integral of the difference between the curves' y -values:

Area= bax+2−x2dx 


The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations:


y=x+2y=x2 


By eliminating y, we obtain an equation for x,


x2=x+2


If we move all x-terms to the left-hand side,


x2−x=2


and complete the square, we obtain


x−212−212=2x−212=49

Taking the root then gives that x=2123. In other words, x=−1 and x=2.

The area of the region is now given by


\displaystyle \begin{align} & \text{Area}=\int\limits_{-1}^{2}{\left( x+2-x^{2} \right)}\,dx=\left[ \frac{x^{2}}{2}+2x-\frac{x^{3}}{3} \right]_{-1}^{2} \\ & =\frac{2^{2}}{2}+2\centerdot 2-\frac{2^{3}}{3}-\left( \frac{\left( -1 \right)^{2}}{2}+2\centerdot \left( -1 \right)-\frac{\left( -1 \right)^{3}}{3} \right) \\ & =2+4-\frac{8}{3}-\frac{1}{2}+2-\frac{1}{3} \\ & =\frac{9}{2} \\ \end{align}

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