Für diese Aufgabe benötigst Du folgendes Grundwissen: partielle Integration
A g e s a m t = A 1 + A 2 A_{gesamt}=A_1+A_2 A g es am t = A 1 + A 2
Zur Flächenberechnung benötigst du eine Stammfunktion G ( x ) G(x) G ( x ) g ( x ) = − x 2 ⋅ x + 2 g(x)=-\frac{x}{2}\cdot\sqrt{x+2} g ( x ) = − 2 x ⋅ x + 2
G ( x ) = ∫ ( − x 2 ) ⋅ x + 2 d x = − 1 2 ∫ x ⋅ x + 2 d x G(x)=\int_{}^{}\left(-\frac{x}{2}\right)\cdot\sqrt{x+2}\; \mathrm{d}x=-\frac{1}{2}\int_{}^{}x\cdot\sqrt{x+2}\; \mathrm{d}x G ( x ) = ∫ ( − 2 x ) ⋅ x + 2 d x = − 2 1 ∫ x ⋅ x + 2 d x
Das Integral kann mit partieller Integration berechnet werden. (Der Vorfaktor wird zunächst nicht berücksichtigt.)
Die partielle Integration berechnet mit folgender Formel das Integral über ein Produkt von zwei Funktionen: ∫ u ′ ( x ) ⋅ v ( x ) d x = u ( x ) ⋅ v ( x ) − ∫ u ( x ) ⋅ v ′ ( x ) d x \int_{}^{}u'(x)\cdot v(x)\; \mathrm{d}x=u(x)\cdot v(x)-\int_{}^{}u(x)\cdot v'(x)\; \mathrm{d}x ∫ u ′ ( x ) ⋅ v ( x ) d x = u ( x ) ⋅ v ( x ) − ∫ u ( x ) ⋅ v ′ ( x ) d x
Es soll das Integral ∫ x + 2 ⋅ x d x \int_{}^{}\sqrt{x+2}\cdot x\; \mathrm{d}x ∫ x + 2 ⋅ x d x
Setze dazu u ′ ( x ) = x + 2 u{'}(x)=\sqrt{x+2} u ′ ( x ) = x + 2 v ( x ) = x v(x)=x v ( x ) = x
Berechne nun u ( x ) u(x) u ( x ) x + 2 \sqrt{x+2} x + 2
u ′ ( x ) \displaystyle u'(x) u ′ ( x ) = = = x + 2 \displaystyle \sqrt{x+2} x + 2 ↓
Schreibe die Wurzel als Potenz.
= = = ( x + 2 ) 1 2 \displaystyle (x+2)^\frac{1}{2} ( x + 2 ) 2 1 ↓
Integriere über x x x
u ( x ) \displaystyle u(x) u ( x ) = = = ∫ ( x + 2 ) 1 2 d x \displaystyle \int_{}^{}(x+2)^\frac{^1}{2}\; \mathrm{d}x ∫ ( x + 2 ) 2 1 d x ↓
Wende die Regel für Potenzfunktionen an.
= = = ( x + 2 ) 1 2 + 1 1 2 + 1 \displaystyle \dfrac{(x+2)^{\frac{1}{2}+1}}{{\frac{1}{2}+1}} 2 1 + 1 ( x + 2 ) 2 1 + 1 ↓
Vereinfache.
= = = ( x + 2 ) 3 2 3 2 \displaystyle \dfrac{(x+2)^{\frac{3}{2}}}{\frac{3}{2}} 2 3 ( x + 2 ) 2 3 = = = 2 3 ⋅ ( x + 2 ) 3 2 \displaystyle \frac{2}{3}\cdot(x+2)^{\frac{3}{2}} 3 2 ⋅ ( x + 2 ) 2 3
Somit ist u ( x ) = 2 3 ⋅ ( x + 2 ) 3 2 u(x)= \frac{2}{3}\cdot(x+2)^{\frac{3}{2}} u ( x ) = 3 2 ⋅ ( x + 2 ) 2 3 v ′ ( x ) = 1 v'(x)=1 v ′ ( x ) = 1
Nun kann die partielle Integration durchgeführt werden.
∫ u ′ ( x ) ⋅ v ( x ) d x \displaystyle \int_{}^{}u'(x)\cdot v(x)\; \mathrm{d}x ∫ u ′ ( x ) ⋅ v ( x ) d x = = = u ( x ) ⋅ v ( x ) − ∫ u ( x ) ⋅ v ′ ( x ) d x \displaystyle u(x)\cdot v(x)-\int_{}^{}u(x)\cdot v'(x)\; \mathrm{d}x u ( x ) ⋅ v ( x ) − ∫ u ( x ) ⋅ v ′ ( x ) d x ↓
Setze u ( x ) = 2 3 ⋅ ( x + 2 ) 3 2 u(x)=\frac{2}{3}\cdot(x+2)^{\frac{3}{2}} u ( x ) = 3 2 ⋅ ( x + 2 ) 2 3 v ( x ) = x v(x)=x v ( x ) = x v ′ ( x ) = 1 v'(x)=1 v ′ ( x ) = 1
∫ x + 2 ⋅ x d x \displaystyle \int_{}^{}\sqrt{x+2}\cdot x\; \mathrm{d}x ∫ x + 2 ⋅ x d x = = = 2 3 ⋅ ( x + 2 ) 3 2 ⋅ x − ∫ 2 3 ⋅ ( x + 2 ) 3 2 ⋅ 1 d x \displaystyle \frac{2}{3}\cdot(x+2)^{\frac{3}{2}}\cdot x-\int_{}^{}\frac{2}{3}\cdot(x+2)^{\frac{3}{2}}\cdot 1\; \mathrm{d}x 3 2 ⋅ ( x + 2 ) 2 3 ⋅ x − ∫ 3 2 ⋅ ( x + 2 ) 2 3 ⋅ 1 d x
Berechne nun das hintere Integral durch Anwendung der Regel für Potenzfunktionen :
∫ 2 3 ⋅ ( x + 2 ) 3 2 ⋅ 1 d x = 2 3 ∫ ( x + 2 ) 3 2 d x \int_{}^{}\frac{2}{3}\cdot(x+2)^{\frac{3}{2}}\cdot 1\; \mathrm{d}x=\frac{2}{3}\int_{}^{}(x+2)^{\frac{3}{2}}\; \mathrm{d}x ∫ 3 2 ⋅ ( x + 2 ) 2 3 ⋅ 1 d x = 3 2 ∫ ( x + 2 ) 2 3 d x
2 3 ∫ ( x + 2 ) 3 2 d x \displaystyle \frac{2}{3}\int_{}^{}(x+2)^{\frac{3}{2}}\; \mathrm{d}x 3 2 ∫ ( x + 2 ) 2 3 d x = = = 2 3 ( ( x + 2 ) 3 2 + 1 3 2 + 1 ) \displaystyle \frac{2}{3}\left(\dfrac{(x+2)^{\frac{3}{2}+1}}{{\frac{3}{2}+1}}\right) 3 2 ( 2 3 + 1 ( x + 2 ) 2 3 + 1 ) ↓
Vereinfache.
= = = 2 3 ( ( x + 2 ) 5 2 5 2 ) \displaystyle \frac{2}{3}\left(\dfrac{(x+2)^{\frac{5}{2}}}{\frac{5}{2}}\right) 3 2 ( 2 5 ( x + 2 ) 2 5 ) ↓
Vereinfache.
= = = 4 15 ( x + 2 ) 5 2 \displaystyle \frac{4}{15}(x+2)^{\frac{5}{2}} 15 4 ( x + 2 ) 2 5
Damit folgt für die Berechnung von ∫ x + 2 ⋅ x d x \int_{}^{}\sqrt{x+2}\cdot x\; \mathrm{d}x ∫ x + 2 ⋅ x d x
∫ x + 2 ⋅ x d x \displaystyle \int_{}^{}\sqrt{x+2}\cdot x\; \mathrm{d}x ∫ x + 2 ⋅ x d x = = = 2 3 ⋅ ( x + 2 ) 3 2 ⋅ x − 4 15 ( x + 2 ) 5 2 \displaystyle \frac{2}{3}\cdot(x+2)^{\frac{3}{2}}\cdot x-\frac{4}{15}(x+2)^{\frac{5}{2}} 3 2 ⋅ ( x + 2 ) 2 3 ⋅ x − 15 4 ( x + 2 ) 2 5 ↓
Vereinfache die rechte Seite durch Ausklammern von ( x + 2 ) 3 2 (x+2)^{\frac{3}{2}} ( x + 2 ) 2 3
= = = ( x + 2 ) 3 2 ⋅ ( 2 3 ⋅ x − 4 15 ⋅ ( x + 2 ) ) \displaystyle (x+2)^{\frac{3}{2}}\cdot\left(\frac{2}{3}\cdot x-\frac{4}{15}\cdot(x+2)\right) ( x + 2 ) 2 3 ⋅ ( 3 2 ⋅ x − 15 4 ⋅ ( x + 2 ) ) ↓
Vereinfache die hintere Klammer.
= = = ( x + 2 ) 3 2 ⋅ ( 2 3 ⋅ x − 4 15 ⋅ x − 8 15 ) \displaystyle (x+2)^{\frac{3}{2}}\cdot\left(\frac{2}{3}\cdot x-\frac{4}{15}\cdot x-\frac{8}{15}\right) ( x + 2 ) 2 3 ⋅ ( 3 2 ⋅ x − 15 4 ⋅ x − 15 8 ) = = = ( x + 2 ) 3 2 ⋅ ( 10 15 ⋅ x − 4 15 ⋅ x − 8 15 ) \displaystyle (x+2)^{\frac{3}{2}}\cdot\left(\frac{10}{15}\cdot x-\frac{4}{15}\cdot x-\frac{8}{15}\right) ( x + 2 ) 2 3 ⋅ ( 15 10 ⋅ x − 15 4 ⋅ x − 15 8 ) = = = ( x + 2 ) 3 2 ⋅ ( 6 15 ⋅ x − 8 15 ) \displaystyle (x+2)^{\frac{3}{2}}\cdot\left(\frac{6}{15}\cdot x-\frac{8}{15}\right) ( x + 2 ) 2 3 ⋅ ( 15 6 ⋅ x − 15 8 ) ↓
Klammere 2 15 \frac{2}{15} 15 2
= = = 2 15 ⋅ ( x + 2 ) 3 2 ⋅ ( 3 x − 4 ) \displaystyle \frac{2}{15}\cdot(x+2)^{\frac{3}{2}}\cdot(3x-4) 15 2 ⋅ ( x + 2 ) 2 3 ⋅ ( 3 x − 4 )
Eine Stammfunktion lautet somit:
G ( x ) = − 1 2 ∫ x ⋅ x + 2 d x = − 1 2 ⋅ 2 15 ⋅ ( x + 2 ) 3 2 ⋅ ( 3 x − 4 ) G(x)=-\frac{1}{2}\int_{}^{}x\cdot\sqrt{x+2}\; \mathrm{d}x=-\frac{1}{2}\cdot\frac{2}{15}\cdot(x+2)^{\frac{3}{2}}\cdot(3x-4) G ( x ) = − 2 1 ∫ x ⋅ x + 2 d x = − 2 1 ⋅ 15 2 ⋅ ( x + 2 ) 2 3 ⋅ ( 3 x − 4 )
G ( x ) = − 1 15 ⋅ ( x + 2 ) 3 2 ⋅ ( 3 x − 4 ) \displaystyle G(x)=-\frac{1}{15}\cdot(x+2)^{\frac{3}{2}}\cdot(3x-4) G ( x ) = − 15 1 ⋅ ( x + 2 ) 2 3 ⋅ ( 3 x − 4 ) Berechne nun die Fläche A 1 A_1 A 1
Die Integrationsgrenzen werden der Abbildung entnommen. Die untere Grenze ist x = − 2 x=-2 x = − 2 x = 0 x=0 x = 0
A 1 = 2 ⋅ ∫ − 2 0 x ⋅ x + 2 d x A_1=2\cdot\displaystyle\int_{-2}^{0}x\cdot\sqrt{x+2}\; \mathrm{d}x A 1 = 2 ⋅ ∫ − 2 0 x ⋅ x + 2 d x
∫ − 2 0 x ⋅ x + 2 d x \displaystyle \displaystyle\int_{-2}^{0}x\cdot\sqrt{x+2}\; \mathrm{d}x ∫ − 2 0 x ⋅ x + 2 d x = = = [ − 1 15 ⋅ ( x + 2 ) 3 2 ⋅ ( 3 x − 4 ) ] − 2 0 \displaystyle \left[-\frac{1}{15}\cdot(x+2)^{\frac{3}{2}}\cdot(3x-4)\right]_{-2}^{0} [ − 15 1 ⋅ ( x + 2 ) 2 3 ⋅ ( 3 x − 4 ) ] − 2 0 ↓
Setze die Grenzen ein.
= = = ( − 1 15 ⋅ ( 0 + 2 ) 3 2 ⋅ ( 3 ⋅ 0 − 4 ) ) − ( − 1 15 ⋅ ( − 2 + 2 ) 3 2 ⋅ ( 3 ⋅ ( − 2 ) − 4 ) ) \displaystyle \left(-\frac{1}{15}\cdot(0+2)^{\frac{3}{2}}\cdot(3\cdot 0-4)\right)-\left(-\frac{1}{15}\cdot(-2+2)^{\frac{3}{2}}\cdot(3\cdot (-2)-4)\right) ( − 15 1 ⋅ ( 0 + 2 ) 2 3 ⋅ ( 3 ⋅ 0 − 4 ) ) − ( − 15 1 ⋅ ( − 2 + 2 ) 2 3 ⋅ ( 3 ⋅ ( − 2 ) − 4 ) ) ↓
Vereinfache.
= = = ( − 1 15 ⋅ 2 3 2 ⋅ ( − 4 ) ) − ( − 1 15 ⋅ 0 3 2 ⋅ ( − 6 − 4 ) ) \displaystyle \left(-\frac{1}{15}\cdot2^{\frac{3}{2}}\cdot(-4)\right)-\left(-\frac{1}{15}\cdot0^{\frac{3}{2}}\cdot(-6-4)\right) ( − 15 1 ⋅ 2 2 3 ⋅ ( − 4 ) ) − ( − 15 1 ⋅ 0 2 3 ⋅ ( − 6 − 4 ) ) ↓
Vereinfache.
Wegen 0 3 2 = 0 0^\frac{^3}{2}=0 0 2 3 = 0
2 3 2 = 2 ⋅ 2 2^\frac{^3}{2}=2\cdot\sqrt{2} 2 2 3 = 2 ⋅ 2
= = = 4 15 ⋅ 2 ⋅ 2 \displaystyle \frac{4}{15}\cdot2\cdot\sqrt{2} 15 4 ⋅ 2 ⋅ 2 = = = 8 15 ⋅ 2 \displaystyle \frac{8}{15}\cdot\sqrt{2} 15 8 ⋅ 2
Die Fläche A 1 A_1 A 1 2 ⋅ 8 15 ⋅ 2 = 16 15 ⋅ 2 m 2 2\cdot\frac{8}{15}\cdot\sqrt{2}=\frac{16}{15}\cdot\sqrt{2}\;\mathrm{m^2} 2 ⋅ 15 8 ⋅ 2 = 15 16 ⋅ 2 m 2
Berechne nun die Fläche A 2 A_2 A 2
Die Integrationsgrenzen werden der Abbildung entnommen. Die untere Grenze ist x = 0 x=0 x = 0 x = 1 x=1 x = 1 g ( x ) g(x) g ( x )
A 2 = 2 ⋅ ∣ ∫ 0 1 x ⋅ x + 2 d x ∣ A_2=2\cdot\left|\displaystyle\int_{0}^{1}x\cdot\sqrt{x+2}\; \mathrm{d}x\right| A 2 = 2 ⋅ ∫ 0 1 x ⋅ x + 2 d x
∫ 0 1 x ⋅ x + 2 d x \displaystyle \displaystyle\int_{0}^{1}x\cdot\sqrt{x+2}\; \mathrm{d}x ∫ 0 1 x ⋅ x + 2 d x = = = [ − 1 15 ⋅ ( x + 2 ) 3 2 ⋅ ( 3 x − 4 ) ] 0 1 \displaystyle \left[-\frac{1}{15}\cdot(x+2)^{\frac{3}{2}}\cdot(3x-4)\right]_{0}^{1} [ − 15 1 ⋅ ( x + 2 ) 2 3 ⋅ ( 3 x − 4 ) ] 0 1 ↓
Setze die Grenzen ein.
= = = ( − 1 15 ⋅ ( 1 + 2 ) 3 2 ⋅ ( 3 ⋅ 1 − 4 ) ) − ( − 1 15 ⋅ ( 0 + 2 ) 3 2 ⋅ ( 3 ⋅ 0 − 4 ) ) \displaystyle \left(-\frac{1}{15}\cdot(1+2)^{\frac{3}{2}}\cdot(3\cdot 1-4)\right)-\left(-\frac{1}{15}\cdot(0+2)^{\frac{3}{2}}\cdot(3\cdot 0-4)\right) ( − 15 1 ⋅ ( 1 + 2 ) 2 3 ⋅ ( 3 ⋅ 1 − 4 ) ) − ( − 15 1 ⋅ ( 0 + 2 ) 2 3 ⋅ ( 3 ⋅ 0 − 4 ) ) ↓
Vereinfache.
= = = ( − 1 15 ⋅ 3 3 2 ⋅ ( − 1 ) ) − ( − 1 15 ⋅ ( 2 ) 3 2 ⋅ ( − 4 ) ) \displaystyle \left(-\frac{1}{15}\cdot3^{\frac{3}{2}}\cdot(-1)\right)-\left(-\frac{1}{15}\cdot(2)^{\frac{3}{2}}\cdot(-4)\right) ( − 15 1 ⋅ 3 2 3 ⋅ ( − 1 ) ) − ( − 15 1 ⋅ ( 2 ) 2 3 ⋅ ( − 4 ) ) ↓
Vereinfache.
3 3 2 = 3 ⋅ 3 3^\frac{^3}{2}=3\cdot\sqrt{3} 3 2 3 = 3 ⋅ 3 2 3 2 = 2 ⋅ 2 2^\frac{^3}{2}=2\cdot\sqrt{2} 2 2 3 = 2 ⋅ 2
= = = ( 1 15 ⋅ 3 ⋅ 3 ) − ( 4 15 ⋅ 2 ⋅ 2 ) \displaystyle \left(\frac{1}{15}\cdot3\cdot\sqrt{3}\right)-\left(\frac{4}{15}\cdot2\cdot\sqrt{2}\right) ( 15 1 ⋅ 3 ⋅ 3 ) − ( 15 4 ⋅ 2 ⋅ 2 ) ↓
Kürze und vereinfache.
= = = 1 5 ⋅ 3 − 8 15 ⋅ 2 \displaystyle \frac{1}{5}\cdot\sqrt{3}-\frac{8}{15}\cdot\sqrt{2} 5 1 ⋅ 3 − 15 8 ⋅ 2
Die Fläche A 2 A_2 A 2
A 2 = 2 ⋅ ∣ ∫ 0 1 x ⋅ x + 2 d x ∣ A_2=2\cdot\left|\displaystyle\int_{0}^{1}x\cdot\sqrt{x+2}\; \mathrm{d}x\right| A 2 = 2 ⋅ ∫ 0 1 x ⋅ x + 2 d x ⇒ \;\Rightarrow\; ⇒
A 2 = 2 ⋅ ∣ 1 5 ⋅ 3 − 8 15 ⋅ 2 ∣ A_2=2\cdot\left|\frac{1}{5}\cdot\sqrt{3}-\frac{8}{15}\cdot\sqrt{2}\right| A 2 = 2 ⋅ 5 1 ⋅ 3 − 15 8 ⋅ 2
∣ 1 5 ⋅ 3 − 8 15 ⋅ 2 ∣ = − ( 1 5 ⋅ 3 − 8 15 ⋅ 2 ) = ( 8 15 ⋅ 2 − 1 5 ⋅ 3 ) \left|\frac{1}{5}\cdot\sqrt{3}-\frac{8}{15}\cdot\sqrt{2}\right|=-\left(\frac{1}{5}\cdot\sqrt{3}-\frac{8}{15}\cdot\sqrt{2}\right)=\left(\frac{8}{15}\cdot\sqrt{2}-\frac{1}{5}\cdot\sqrt{3}\right) 5 1 ⋅ 3 − 15 8 ⋅ 2 = − ( 5 1 ⋅ 3 − 15 8 ⋅ 2 ) = ( 15 8 ⋅ 2 − 5 1 ⋅ 3 )
A 2 = 2 ⋅ ( 8 15 ⋅ 2 − 1 5 ⋅ 3 ) = 16 15 ⋅ 2 − 2 5 ⋅ 3 A_2=2\cdot\left(\frac{8}{15}\cdot\sqrt{2}-\frac{1}{5}\cdot\sqrt{3}\right)=\frac{16}{15}\cdot\sqrt{2}-\frac{2}{5}\cdot\sqrt{3} A 2 = 2 ⋅ ( 15 8 ⋅ 2 − 5 1 ⋅ 3 ) = 15 16 ⋅ 2 − 5 2 ⋅ 3
Die Fläche A 2 A_2 A 2 ( 16 15 ⋅ 2 − 2 5 ⋅ 3 ) m 2 \left(\frac{16}{15}\cdot\sqrt{2}-\frac{2}{5}\cdot\sqrt{3}\right)\;\mathrm{m^2} ( 15 16 ⋅ 2 − 5 2 ⋅ 3 ) m 2
A g e s a m t = A 1 + A 2 = 16 15 ⋅ 2 + 16 15 ⋅ 2 − 2 5 ⋅ 3 A_{gesamt}=A_1+A_2=\frac{16}{15}\cdot\sqrt{2}+\frac{16}{15}\cdot\sqrt{2}-\frac{2}{5}\cdot\sqrt{3} A g es am t = A 1 + A 2 = 15 16 ⋅ 2 + 15 16 ⋅ 2 − 5 2 ⋅ 3
A g e s a m t = 32 15 ⋅ 2 − 2 5 ⋅ 3 A_{gesamt}=\frac{32}{15}\cdot\sqrt{2}-\frac{2}{5}\cdot\sqrt{3} A g es am t = 15 32 ⋅ 2 − 5 2 ⋅ 3
A g e s a m t ≈ 2 , 32 m 2 A_{gesamt}\approx2{,}32\;\mathrm{m^2} A g es am t ≈ 2 , 32 m 2
Der Maler muss eine Fläche von etwa 2 , 32 m 2 2{,}32\;\mathrm{m^2} 2 , 32 m 2
