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Lösung 2.1:4b

Aus Online Mathematik Brückenkurs 2

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By completing the square of the equation of the curve


y=x2+2x+2=x22x2=x12122=x12+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= bax2+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=x12+3 

or


x12=3 .

Taking the root gives x=13 . The points of intersection x=13  and x=1+3 .

The area we are looking for is therefore given by

Area =1+313x2+2x+2dx 


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

Area= =1+313x12+3dx 


which seems easier. Because the expression x1 inside the square is a linear expression, we can write down a primitive function “in the usual way”,

Area =3x13+3x1+313 


(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,


Area=31+313+31+331313+313=333+3+33+3333+33=3333+33+3333+33=333+33333+33=3+333+33=1+31+33=43


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


=1+313x2+2x+2dx=