Lösung 2.1:1b

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The graph of the function
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The graph of the function <math>y=2x+1</math> is a straight line which cuts the ''y''-axis at <math>y=1</math> and has slope 2.
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<math>y=2x+1</math>
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is a straight line which cuts the
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<math>y</math>
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-axis at
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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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The integral's value is the area under the straight line and between <math>x=0</math>
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and <math>x=1</math>.
[[Image:2_1_1_b1.gif|center]]
[[Image:2_1_1_b1.gif|center]]
We can divide up the region under the graph into a square and rectangle,
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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The value of the integral is
The value of the integral is
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{{Displayed math||<math>\begin{align}
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<math>\int\limits_{0}^{1}{\left( 2x+1 \right)\,}dx=</math>
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\int\limits_{0}^{1} (2x+1)\,dx
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(area of the square) + (area of the triangle)
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&= \text{(area of the square)} + \text{(area of the triangle)}\\
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<math>1\centerdot 1+\frac{1}{2}\centerdot 1\centerdot 2=2</math>
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&= 1\cdot 1 + \frac{1}{2}\cdot 1\cdot 2 = 2\,\textrm{.}
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\end{align}</math>}}

Version vom 12:09, 21. Okt. 2008

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

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

\int\limits_{0}^{1} (2x+1)\,dx &= \text{(area of the square)} + \text{(area of the triangle)}\\ &= 1\cdot 1 + \frac{1}{2}\cdot 1\cdot 2 = 2\,\textrm{.} \end{align}

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