fläche y=x,4x+y^2=12

Lösung

fläche

y = x, 4 x + y^{2} = 12

\frac{64}{3}

Dezimale

21.33333 \dots

Schritte zur Lösung

Stelle

y

nach

4 x + y^{2} = 12

um:

y = 2 3 - x, y = - 2 3 - x

f_{1} (x) = x

f_{2} (x) = 2 3 - x

Um die Schnittpunkte zu finden, löse

f_{i} (x) = f_{j} (x)

x = 2 3 - x : x = 2

x = - 2 3 - x : x = - 6

2 3 - x = - 2 3 - x : x = 3

= \int_{- 6}^{2} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{2}^{3} ∣ f_{3} (x) - f_{2} (x) ∣ d x

= \int_{- 6}^{2} x - (- 2 3 - x) d x + \int_{2}^{3} - 2 3 - x - 2 3 - x d x

\int_{- 6}^{2} x - (- 2 3 - x) d x = \frac{56}{3}

\int_{2}^{3} - 2 3 - x - 2 3 - x d x = \frac{8}{3}

= \frac{56}{3} + \frac{8}{3}

Vereinfache

= \frac{64}{3}

in v erse l a pl a ce {\frac{5}{s ^{2} + 49}}

d er i v a t i v e (a + 3 y^{2} + b)^{\frac{1}{2}}

d er i v a t i v e f (x) = \frac{π ^{2}}{x ^{2} + 4 x}

\int - \frac{1}{27} sin (3 x) d x

\int \frac{3 - 5 x}{9 - 5 x ^{2}} d x