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

Lösung 2.1:4b

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.1:4b moved to Solution 2.1:4b: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
By completing the square of the equation of the curve
-
<center> [[Image:2_1_4b-1(4).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
-
{{NAVCONTENT_START}}
+
<math>\begin{align}
-
<center> [[Image:2_1_4b-2(4).gif]] </center>
+
& y=-x^{2}+2x+2=-\left( x^{2}-2x-2 \right) \\
-
{{NAVCONTENT_STOP}}
+
& =-\left( \left( x-1 \right)^{2}-1^{2}-2 \right)=-\left( x-1 \right)^{2}+3 \\
-
{{NAVCONTENT_START}}
+
\end{align}</math>
-
<center> [[Image:2_1_4b-3(4).gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
-
{{NAVCONTENT_START}}
+
we can read off that the curve is a downward parabola with maximum value
-
<center> [[Image:2_1_4b-4(4).gif]] </center>
+
<math>y=\text{3 }</math>
-
{{NAVCONTENT_STOP}}
+
when
 +
<math>x=\text{1}</math>
 +
 
[[Image:2_1_4_b.gif|center]]
[[Image:2_1_4_b.gif|center]]
 +
 +
 +
The region whose area we shall determine is the one shaded in the figure.
 +
 +
We can express this area using the integral
 +
 +
Area=
 +
<math>\int\limits_{a}^{b}{\left( -x^{2}+2x+2 \right)}\,dx</math>
 +
 +
where
 +
<math>a</math>
 +
and
 +
<math>b</math>
 +
are the
 +
<math>x</math>
 +
-coordinates for the points of intersection between the parabola and the
 +
<math>x</math>
 +
-axis.
 +
 +
A solution plan is to first determine the intersection points,
 +
<math>x=a</math>
 +
and
 +
<math>x=b</math>, and then calculate the area using the integral formula above.
 +
 +
The parabola cuts the
 +
<math>x</math>
 +
-axis when its
 +
<math>y</math>
 +
-coordinate is zero, i.e.
 +
 +
 +
<math>0=-x^{2}+2x+2</math>
 +
 +
 +
and because we have already completed the square of the right-hand side once, the equation can be written as
 +
 +
 +
<math>0=-\left( x-1 \right)^{2}+3</math>
 +
 +
or
 +
 +
 +
<math>\left( x-1 \right)^{2}=3</math>.
 +
 +
Taking the root gives
 +
<math>x=1\pm \sqrt{3}</math>. The points of intersection
 +
<math>x=1-\sqrt{3}</math>
 +
and
 +
<math>x=1+\sqrt{3}</math>.
 +
 +
The area we are looking for is therefore given by
 +
 +
Area
 +
<math>=\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -x^{2}+2x+2 \right)}\,dx</math>
 +
 +
 +
Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,
 +
 +
Area=
 +
<math>=\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -\left( x-1 \right)^{2}+3 \right)}\,dx</math>
 +
 +
 +
which seems easier. Because the expression
 +
<math>x-1</math>
 +
inside the square is a linear expression, we can write down a primitive function “in the usual way”,
 +
 +
Area
 +
<math>=\left[ -\frac{\left( x-1 \right)^{3}}{3}+3x \right]_{1-\sqrt{3}}^{1+\sqrt{3}}</math>
 +
 +
 +
(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integral back). Hence,
 +
 +
 +
<math>\begin{align}
 +
& \text{Area}=-\frac{\left( 1+\sqrt{3}-1 \right)^{3}}{3}+3\left( 1+\sqrt{3} \right)-\left( -\frac{\left( 1-\sqrt{3}-1 \right)^{3}}{3}+3\left( 1-\sqrt{3} \right) \right) \\
 +
& =-\frac{\left( \sqrt{3} \right)^{3}}{3}+3+3\sqrt{3}+\frac{\left( -\sqrt{3} \right)^{3}}{3}-3+3\sqrt{3} \\
 +
& =-\frac{\sqrt{3}\sqrt{3}\sqrt{3}}{3}+3\sqrt{3}+\frac{\left( -\sqrt{3} \right)\left( -\sqrt{3} \right)\left( -\sqrt{3} \right)}{3}+3\sqrt{3} \\
 +
& =-\frac{3\sqrt{3}}{3}+3\sqrt{3}-\frac{3\sqrt{3}}{3}+3\sqrt{3} \\
 +
& =-\sqrt{3}+3\sqrt{3}-\sqrt{3}+3\sqrt{3} \\
 +
& =\left( -1+3-1+3 \right)\sqrt{3}=4\sqrt{3} \\
 +
\end{align}</math>
 +
 +
 +
NOTE: The calculations become a lot more complicated if one starts from
 +
 +
 +
<math>=\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -x^{2}+2x+2 \right)}\,dx=....</math>

Version vom 11:32, 18. Okt. 2008

By completing the square of the equation of the curve


y=−x2+2x+2=−x2−2x−2=−x−12−12−2=−x−12+3 


we can read off that the curve is a downward parabola with maximum value y=3 when x=1



The region whose area we shall determine is the one shaded in the figure.

We can express this area using the integral

Area= ba−x2+2x+2dx 

where a and b are the x -coordinates for the points of intersection between the parabola and the x -axis.

A solution plan is to first determine the intersection points, x=a and x=b, and then calculate the area using the integral formula above.

The parabola cuts the x -axis when its y -coordinate is zero, i.e.


0=−x2+2x+2


and because we have already completed the square of the right-hand side once, the equation can be written as


0=−x−12+3 

or


x−12=3 .

Taking the root gives x=13 . The points of intersection x=1−3  and x=1+3 .

The area we are looking for is therefore given by

Area =1+31−3−x2+2x+2dx 


Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,

Area= =1+31−3−x−12+3dx 


which seems easier. Because the expression x−1 inside the square is a linear expression, we can write down a primitive function “in the usual way”,

Area =−3x−13+3x1+31−3 


(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integral back). Hence,


Area=−31+3−13+31+3−−31−3−13+31−3=−333+3+33+3−33−3+33=−3333+33+3−3−3−3+33=−333+33−333+33=−3+33−3+33=−1+3−1+33=43


NOTE: The calculations become a lot more complicated if one starts from


=1+31−3−x2+2x+2dx= 

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