Lösung 2.1:4b

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


\displaystyle \begin{align} & y=-x^{2}+2x+2=-\left( x^{2}-2x-2 \right) \\ & =-\left( \left( x-1 \right)^{2}-1^{2}-2 \right)=-\left( x-1 \right)^{2}+3 \\ \end{align}


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



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

We can express this area using the integral

Area= \displaystyle \int\limits_{a}^{b}{\left( -x^{2}+2x+2 \right)}\,dx

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

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

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


\displaystyle 0=-x^{2}+2x+2


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


\displaystyle 0=-\left( x-1 \right)^{2}+3

or


\displaystyle \left( x-1 \right)^{2}=3.

Taking the root gives \displaystyle x=1\pm \sqrt{3}. The points of intersection \displaystyle x=1-\sqrt{3} and \displaystyle x=1+\sqrt{3}.

The area we are looking for is therefore given by

Area \displaystyle =\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -x^{2}+2x+2 \right)}\,dx


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

Area= \displaystyle =\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -\left( x-1 \right)^{2}+3 \right)}\,dx


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

Area \displaystyle =\left[ -\frac{\left( x-1 \right)^{3}}{3}+3x \right]_{1-\sqrt{3}}^{1+\sqrt{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,


\displaystyle \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}


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


\displaystyle =\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -x^{2}+2x+2 \right)}\,dx=....

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