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Lösung 2.1:4b

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
By completing the square of the equation of the curve
By completing the square of the equation of the curve
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
y &= -x^2 + 2x + 2\\[5pt]
y &= -x^2 + 2x + 2\\[5pt]
&= -\bigl(x^2 - 2x- 2\bigr)\\[5pt]
&= -\bigl(x^2 - 2x- 2\bigr)\\[5pt]
Zeile 16: Zeile 16:
We can express this area using the integral
We can express this area using the integral
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{{Displayed math||<math>\text{Area} = \int\limits_a^b \bigl(-x^2+2x+2\bigr)\,dx\,,</math>}}
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{{Abgesetzte Formel||<math>\text{Area} = \int\limits_a^b \bigl(-x^2+2x+2\bigr)\,dx\,,</math>}}
where ''a'' and ''b'' are the ''x''-coordinates for the points of intersection between the parabola and the ''x''-axis.
where ''a'' and ''b'' are the ''x''-coordinates for the points of intersection between the parabola and the ''x''-axis.
Zeile 25: Zeile 25:
The parabola cuts the ''x''-axis when its ''y''-coordinate is zero, i.e.
The parabola cuts the ''x''-axis when its ''y''-coordinate is zero, i.e.
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{{Displayed math||<math>0=-x^{2}+2x+2</math>}}
+
{{Abgesetzte Formel||<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
and because we have already completed the square of the right-hand side once, the equation can be written as
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{{Displayed math||<math>0=-(x-1)^2+3</math>}}
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{{Abgesetzte Formel||<math>0=-(x-1)^2+3</math>}}
or
or
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{{Displayed math||<math>(x-1)^2=3\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>(x-1)^2=3\,\textrm{.}</math>}}
Taking the square root gives <math>x = 1\pm \sqrt{3}\,</math>. The points of intersection are <math>x=1-\sqrt{3}</math> and <math>x=1+\sqrt{3}\,</math>.
Taking the square root gives <math>x = 1\pm \sqrt{3}\,</math>. The points of intersection are <math>x=1-\sqrt{3}</math> and <math>x=1+\sqrt{3}\,</math>.
Zeile 39: Zeile 39:
The area we are looking for is therefore given by
The area we are looking for is therefore given by
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{{Displayed math||<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}</math>}}
Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,
Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,
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{{Displayed math||<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,</math>}}
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{{Abgesetzte Formel||<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,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”,
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”,
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{{Displayed math||<math>\text{Area} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\text{Area} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}</math>}}
(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integrand back). Hence,
(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integrand back). Hence,
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\text{Area} &= -\frac{(1+\sqrt{3}-1)^3}{3}+3(1+\sqrt{3}\,)-\Bigl(-\frac{(1-\sqrt{3}-1)^3}{3}+3(1-\sqrt{3}\,)\Bigr)\\[5pt]
\text{Area} &= -\frac{(1+\sqrt{3}-1)^3}{3}+3(1+\sqrt{3}\,)-\Bigl(-\frac{(1-\sqrt{3}-1)^3}{3}+3(1-\sqrt{3}\,)\Bigr)\\[5pt]
&= -\frac{(\sqrt{3}\,)^3}{3} + 3 + 3\sqrt{3} + \frac{(-\sqrt{3}\,)^3}{3} - 3 + 3\sqrt{3}\\[5pt]
&= -\frac{(\sqrt{3}\,)^3}{3} + 3 + 3\sqrt{3} + \frac{(-\sqrt{3}\,)^3}{3} - 3 + 3\sqrt{3}\\[5pt]
Zeile 64: Zeile 64:
Note: The calculations become a lot more complicated if one starts from
Note: The calculations become a lot more complicated if one starts from
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{{Displayed math||<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}
+
{{Abgesetzte Formel||<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}

Version vom 12:59, 10. Mär. 2009

By completing the square of the equation of the curve

y=−x2+2x+2=−x2−2x−2=−(x−1)2−12−2=−(x−1)2+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−1)2+3

or

(x−1)2=3.

Taking the square root gives x=13 . The points of intersection are 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−1)2+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= −3(x−1)3+3x 1+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 integrand back). Hence,

Area=−3(1+3−1)3+3(1+3)−−3(1−3−1)3+3(1−3)=−3(3)3+3+33+3(−3)3−3+33=−3333+33+3(−3)(−3)(−3)+33=−333+33−333+33=−3+33−3+33=(−1+3−1+3)3=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“