Lösung 2.1:4b
Aus Online Mathematik Brückenkurs 2
By completing the square of the equation of the curve
x2−2x−2
=−
x−1
2−12−2
=−
x−1
2+3
we can read off that the curve is a downward parabola with maximum value
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+2
dx
where
A solution plan is to first determine the intersection points,
The parabola cuts the
and because we have already completed the square of the right-hand side once, the equation can be written as
x−1
2+3
or
x−1
2=3
Taking the root gives
3
3
3
The area we are looking for is therefore given by
Area
1+
31−
3
−x2+2x+2
dx
Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,
Area=
1+
31−
3
−
x−1
2+3
dx
which seems easier. Because the expression
Area
−3
x−1
3+3x
1+
31−
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,
1+
3−1
3+3
1+
3
−
−3
1−
3−1
3+3
1−
3
=−3
3
3+3+3
3+3
−
3
3−3+3
3=−3
3
3
3+3
3+3
−
3
−
3
−
3
+3
3=−33
3+3
3−33
3+3
3=−
3+3
3−
3+3
3=
−1+3−1+3
3=4
3
NOTE: The calculations become a lot more complicated if one starts from
1+
31−
3
−x2+2x+2
dx=