From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	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.
-	<center> [[Image:2_1_1b-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	<center> [[Image:2_1_1b-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
		+	The integral's value is the area under the straight line and between <math>x=0</math>
		+	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,
		+
	[[Image:2_1_1_b2.gif|center]]		[[Image:2_1_1_b2.gif|center]]
		+
		+	and then add up the area to obtain the total area.
		+
		+	The value of the integral is
		+
		+	{{Displayed math||<math>\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}</math>}}

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Current revision

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}