Lösung 2.1:1b

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K (Lösning 2.1:1b moved to Solution 2.1:1b: Robot: moved page)
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{{NAVCONTENT_START}}
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The graph of the function
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<center> [[Image:2_1_1b-1(2).gif]] </center>
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<math>y=2x+1</math>
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{{NAVCONTENT_STOP}}
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is a straight line which cuts the
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{{NAVCONTENT_START}}
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<math>y</math>
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<center> [[Image:2_1_1b-2(2).gif]] </center>
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-axis at
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{{NAVCONTENT_STOP}}
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<math>y=\text{1}</math>
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and has gradient
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<math>2</math>.
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The integral's value is the area under the straight line and between
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<math>x=0\text{ }</math>
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and
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<math>x=\text{1}</math>.
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[[Image:2_1_1_b1.gif|center]]
[[Image:2_1_1_b1.gif|center]]
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We can divide up the region under the graph into a square and rectangle,
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[[Image:2_1_1_b2.gif|center]]
[[Image:2_1_1_b2.gif|center]]
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and then add up the area to obtain the total area.
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The value of the integral is
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<math>\int\limits_{0}^{1}{\left( 2x+1 \right)\,}dx=</math>
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(area of the square) + (area of the triangle)
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<math>1\centerdot 1+\frac{1}{2}\centerdot 1\centerdot 2=2</math>

Version vom 11:37, 17. Okt. 2008

The graph of the function \displaystyle y=2x+1 is a straight line which cuts the \displaystyle y -axis at \displaystyle y=\text{1} and has gradient \displaystyle 2.

The integral's value is the area under the straight line and between \displaystyle x=0\text{ } and \displaystyle x=\text{1}.


We can divide up the region under the graph into a square and rectangle,


and then add up the area to obtain the total area.

The value of the integral is


\displaystyle \int\limits_{0}^{1}{\left( 2x+1 \right)\,}dx= (area of the square) + (area of the triangle) \displaystyle 1\centerdot 1+\frac{1}{2}\centerdot 1\centerdot 2=2

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