fläche y=4x,y= 3/2 x,y=45-x^2

Lösung

fläche

y = 4 x, y = \frac{3}{2} x, y = 45 - x^{2}

\frac{5885}{48}

Dezimale

122.60416 \dots

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

4 x = \frac{3}{2} x : x = 0

4 x = 45 - x^{2} : x = - 9, x = 5

\frac{3}{2} x = 45 - x^{2} : x = - \frac{15}{2}, x = 6

= \int_{- 9}^{- \frac{15}{2}} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{- \frac{15}{2}}^{0} ∣ f_{2} (x) - f_{1} (x) ∣ d x + \int_{0}^{5} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{5}^{6} ∣ f_{2} (x) - f_{3} (x) ∣ d x

= \int_{- 9}^{- \frac{15}{2}} 4 x - (45 - x^{2}) d x + \int_{- \frac{15}{2}}^{0} \frac{3}{2} x - 4 x d x + \int_{0}^{5} 4 x - \frac{3}{2} x d x + \int_{5}^{6} \frac{3}{2} x - (45 - x^{2}) d x

\int_{- 9}^{- \frac{15}{2}} 4 x - (45 - x^{2}) d x = \frac{117}{8}

\int_{- \frac{15}{2}}^{0} \frac{3}{2} x - 4 x d x = \frac{1125}{16}

\int_{0}^{5} 4 x - \frac{3}{2} x d x = \frac{125}{4}

\int_{5}^{6} \frac{3}{2} x - (45 - x^{2}) d x = \frac{77}{12}

= \frac{117}{8} + \frac{1125}{16} + \frac{125}{4} + \frac{77}{12}

Vereinfache

= \frac{5885}{48}

\frac{d}{d x} (3 x^{3} ln (x))

d er i v a t i v e n

\int y^{2} d y

\int_{1}^{2} \frac{3}{t ^{4}} d t

x^{2} y^{^{'}} + 2 x^{3} y = y^{2} (1 + 2 x^{2})