F(intervall2) - F(intervall1) = A
Nein, es gilt:
F(obere Grenze) - F ( untere Grenze) = A
f ( x ) = - 2 x + 3
=> F ( x ) = - x 2 + 3 x
Also:
∫0u - 2 x + 3 dx = 1
<=> [ - x 2 + 3 x ]0u = 1
<=> ( - u 2 + 3 u ) - ( - 0 2 + 3 * 0 ) = 1
<=> - u 2 + 3 u = 1
<=> u 2 - 3 u = - 1
<=> u 2 - 3 u + ( 9 / 4 ) = 5 / 4
<=> ( u - 1,5 ) 2 = 5 / 4
<=> u - 1,5 = ± √ ( 5 / 4 ) = ± ( 1 / 2 ) * √ ( 5 )
<=> u = - ( 1 / 2 ) * √ ( 5 ) + 1,5 = ( 1 / 2 ) * ( 3 - √ ( 5 ) )
oder u = + ( 1 / 2 ) * √ ( 5 ) + 1,5 = ( 1 / 2 ) * ( 3 + √ ( 5 ) )