From Förberedande kurs i matematik 2

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-	{{NAVCONTENT_START}}	+	By completing the square of the equation of the curve
-	<center> [[Image:2_1_4b-1(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
-	{{NAVCONTENT_START}}	+	y &= -x^2 + 2x + 2\\[5pt]
-	<center> [[Image:2_1_4b-2(4).gif]] </center>	+	&= -\bigl(x^2 - 2x- 2\bigr)\\[5pt]
-	{{NAVCONTENT_STOP}}	+	&= -\bigl((x-1)^2 - 1^2 - 2\bigr)\\[5pt]
-	{{NAVCONTENT_START}}	+	&= -(x-1)^2 + 3
-	<center> [[Image:2_1_4b-3(4).gif]] </center>	+	\end{align}</math>}}
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+	we can read off that the curve is a downward parabola with maximum value <math>y=3</math> when <math>x=1</math>.
-	<center> [[Image:2_1_4b-4(4).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	[[Image:2_1_4_b.gif|center]]
		+
		+	The region whose area we shall determine is the one shaded in the figure.
		+
		+	We can express this area using the integral
		+
		+	{{Displayed math||<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.
		+
		+	A solution plan is to first determine the intersection points, <math>x=a</math>
		+	and <math>x=b</math>, and then calculate the area using the integral formula above.
		+
		+	The parabola cuts the ''x''-axis when its ''y''-coordinate is zero, i.e.
		+
		+	{{Displayed math||<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
		+
		+	{{Displayed math||<math>0=-(x-1)^2+3</math>}}
		+
		+	or
		+
		+	{{Displayed math||<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>.
		+
		+	The area we are looking for is therefore given by
		+
		+	{{Displayed math||<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,
		+
		+	{{Displayed math||<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”,
		+
		+	{{Displayed math||<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,
		+
		+	{{Displayed math||<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]
		+	&= -\frac{(\sqrt{3}\,)^3}{3} + 3 + 3\sqrt{3} + \frac{(-\sqrt{3}\,)^3}{3} - 3 + 3\sqrt{3}\\[5pt]
		+	&= -\frac{\sqrt{3}\sqrt{3}\sqrt{3}}{3} + 3\sqrt{3} + \frac{(-\sqrt{3}\,)(-\sqrt{3}\,)(-\sqrt{3}\,)}{3} + 3\sqrt{3}\\[5pt]
		+	&= -\frac{3\sqrt{3}}{3} + 3\sqrt{3} - \frac{3\sqrt{3}}{3} + 3\sqrt{3}\\[5pt]
		+	&= -\sqrt{3} + 3\sqrt{3} - \sqrt{3} + 3\sqrt{3}\\[5pt]
		+	&= (-1+3-1+3)\sqrt{3}\\[5pt]
		+	&= 4\sqrt{3}\,\textrm{.}
		+	\end{align}</math>}}
		+
		+
		+	Note: The calculations become a lot more complicated if one starts from
		+
		+	{{Displayed math||<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}

---

Current revision

By completing the square of the equation of the curve

\displaystyle \begin{align} y &= -x^2 + 2x + 2\\[5pt] &= -\bigl(x^2 - 2x- 2\bigr)\\[5pt] &= -\bigl((x-1)^2 - 1^2 - 2\bigr)\\[5pt] &= -(x-1)^2 + 3 \end{align}

we can read off that the curve is a downward parabola with maximum value \displaystyle y=3 when \displaystyle x=1.

The region whose area we shall determine is the one shaded in the figure.

We can express this area using the integral

\displaystyle \text{Area} = \int\limits_a^b \bigl(-x^2+2x+2\bigr)\,dx\,,

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, \displaystyle x=a and \displaystyle 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.

\displaystyle 0=-x^{2}+2x+2

and because we have already completed the square of the right-hand side once, the equation can be written as

\displaystyle 0=-(x-1)^2+3

or

\displaystyle (x-1)^2=3\,\textrm{.}

Taking the square root gives \displaystyle x = 1\pm \sqrt{3}\,. The points of intersection are \displaystyle x=1-\sqrt{3} and \displaystyle x=1+\sqrt{3}\,.

The area we are looking for is therefore given by

\displaystyle \text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}

Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,

\displaystyle \text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,

which seems easier. Because the expression \displaystyle x-1 inside the square is a linear expression, we can write down a primitive function “in the usual way”,

\displaystyle \text{Area} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}

(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,

\displaystyle \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] &= -\frac{(\sqrt{3}\,)^3}{3} + 3 + 3\sqrt{3} + \frac{(-\sqrt{3}\,)^3}{3} - 3 + 3\sqrt{3}\\[5pt] &= -\frac{\sqrt{3}\sqrt{3}\sqrt{3}}{3} + 3\sqrt{3} + \frac{(-\sqrt{3}\,)(-\sqrt{3}\,)(-\sqrt{3}\,)}{3} + 3\sqrt{3}\\[5pt] &= -\frac{3\sqrt{3}}{3} + 3\sqrt{3} - \frac{3\sqrt{3}}{3} + 3\sqrt{3}\\[5pt] &= -\sqrt{3} + 3\sqrt{3} - \sqrt{3} + 3\sqrt{3}\\[5pt] &= (-1+3-1+3)\sqrt{3}\\[5pt] &= 4\sqrt{3}\,\textrm{.} \end{align}

Note: The calculations become a lot more complicated if one starts from

\displaystyle \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots