Aufgaben zu Integralen

(nach einer Abituraufgabe von 2012)

a) Begründe, dass jede Integralfunktion mindestens eine Nullstelle hat.

b) Gib einen Term für eine Funktion $\sf f$ an, sodass die Integralfunktion $\sf F: x \mapsto \int_{1}^x f(t)\operatorname{d}t$ unendlich viele Nullstellen hat.

Begründe, warum es kein $\sf k\in \mathbb{R}^+$ gibt, das folgende Gleichung erfüllt:

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\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

Berechne die Fläche zwischen der x-Achse und $\sf G_f$ im Bereich von $\sf x= a$ bis $\sf x= b$ .

$\sf f(x)=x^3$ $\sf a=0$ $\sf b=1$

$\sf f(x)=x^3$ $\sf a=1$ $\sf b=2$

$\sf f(x)=-x^2+x$ $\sf a=-1$ $\sf b=0$

Berechne

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

Was kann man über die $\sf f$ sagen, wenn man weiß:

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

Berechne die Integrale: $\sf a(x)=6-\dfrac{1}{24}x^2\ ;\ D_a=\mathbb{R}$

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

Berechne.

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

Stelle $\sf f(x)$ integralfrei dar.

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

.lazyload-placeholder { display: none; }

\displaystyle \sf \int_0^ k (x^2+1)\ {d}x=-1

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