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Lösung 2.1:4e

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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<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' ''y''-values,
<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' ''y''-values,
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{{Displayed math||<math>\text{Area} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\text{Area} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}
The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations
The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations
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{{Displayed math||<math>\left\{\begin{align}
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{{Abgesetzte Formel||<math>\left\{\begin{align}
y &= x+2\,,\\[5pt]
y &= x+2\,,\\[5pt]
y &= x^2\,\textrm{.}
y &= x^2\,\textrm{.}
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By eliminating <math>y</math>, we obtain an equation for <math>x</math>,
By eliminating <math>y</math>, we obtain an equation for <math>x</math>,
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{{Displayed math||<math>x^{2}=x+2\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>x^{2}=x+2\,\textrm{.}</math>}}
If we move all ''x''-terms to the left-hand side,
If we move all ''x''-terms to the left-hand side,
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{{Displayed math||<math>x^2-x=2\,,</math>}}
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{{Abgesetzte Formel||<math>x^2-x=2\,,</math>}}
and complete the square, we obtain
and complete the square, we obtain
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\Bigl(x-\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 &= 2\\[5pt]
\Bigl(x-\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 &= 2\\[5pt]
\Bigl(x-\frac{1}{2}\Bigr)^2 &= \frac{9}{4}\,\textrm{.}
\Bigl(x-\frac{1}{2}\Bigr)^2 &= \frac{9}{4}\,\textrm{.}
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The area of the region is now given by
The area of the region is now given by
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\text{Area}
\text{Area}
&= \int\limits_{-1}^2 \bigl(x+2-x^2\bigr)\,dx\\[5pt]
&= \int\limits_{-1}^2 \bigl(x+2-x^2\bigr)\,dx\\[5pt]

Version vom 13:00, 10. Mär. 2009

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

yy=x+2=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−212−212x−212=2=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

Area=2−1x+2−x2dx= 2x2+2x−3x3 2−1=222+22−323−2(−1)2+2(−1)−3(−1)3=2+4−38−21+2−31=29.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:4e“