Lösung 2.1:4b

Aus Online Mathematik Brückenkurs 2

By completing the square of the equation of the curve

y=−x2+2x+2=−x2−2x−2=−x−12−12−2=−x−12+3 

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

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

We can express this area using the integral

Area= ba−x2+2x+2dx 

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, x=a and 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.

0=−x2+2x+2

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

0=−x−12+3 

or

x−12=3 .

Taking the root gives x=13 . The points of intersection \displaystyle x=1-\sqrt{3} and \displaystyle x=1+\sqrt{3}.

The area we are looking for is therefore given by

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

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

Area= \displaystyle =\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -\left( x-1 \right)^{2}+3 \right)}\,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”,

Area \displaystyle =\left[ -\frac{\left( x-1 \right)^{3}}{3}+3x \right]_{1-\sqrt{3}}^{1+\sqrt{3}}

(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integral back). Hence,

\displaystyle \begin{align} & \text{Area}=-\frac{\left( 1+\sqrt{3}-1 \right)^{3}}{3}+3\left( 1+\sqrt{3} \right)-\left( -\frac{\left( 1-\sqrt{3}-1 \right)^{3}}{3}+3\left( 1-\sqrt{3} \right) \right) \\ & =-\frac{\left( \sqrt{3} \right)^{3}}{3}+3+3\sqrt{3}+\frac{\left( -\sqrt{3} \right)^{3}}{3}-3+3\sqrt{3} \\ & =-\frac{\sqrt{3}\sqrt{3}\sqrt{3}}{3}+3\sqrt{3}+\frac{\left( -\sqrt{3} \right)\left( -\sqrt{3} \right)\left( -\sqrt{3} \right)}{3}+3\sqrt{3} \\ & =-\frac{3\sqrt{3}}{3}+3\sqrt{3}-\frac{3\sqrt{3}}{3}+3\sqrt{3} \\ & =-\sqrt{3}+3\sqrt{3}-\sqrt{3}+3\sqrt{3} \\ & =\left( -1+3-1+3 \right)\sqrt{3}=4\sqrt{3} \\ \end{align}

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

\displaystyle =\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -x^{2}+2x+2 \right)}\,dx=....