fläche y=4x,y=4,y= 4/9 x^24

Lösung

fläche

y = 4 x, y = 4, y = \frac{4}{9} x^{2} 4

\frac{59}{8}

Dezimale

7.375

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

4 x = 4 : x = 1

4 x = \frac{4}{9} x^{2} \cdot 4 : x = \frac{9}{4}, x = 0

4 = \frac{4}{9} x^{2} \cdot 4 : x = \frac{3}{2}, x = - \frac{3}{2}

= \int_{- \frac{3}{2}}^{0} ∣ f_{2} (x) - f_{3} (x) ∣ d x + \int_{0}^{1} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{1}^{\frac{3}{2}} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{\frac{3}{2}}^{\frac{9}{4}} ∣ f_{1} (x) - f_{3} (x) ∣ d x

= \int_{- \frac{3}{2}}^{0} 4 - \frac{4}{9} x^{2} \cdot 4 d x + \int_{0}^{1} ∣ 4 x - 4 ∣ d x + \int_{1}^{\frac{3}{2}} ∣ 4 x - 4 ∣ d x + \int_{\frac{3}{2}}^{\frac{9}{4}} 4 x - \frac{4}{9} x^{2} \cdot 4 d x

\int_{- \frac{3}{2}}^{0} 4 - \frac{4}{9} x^{2} \cdot 4 d x = 4

\int_{0}^{1} ∣ 4 x - 4 ∣ d x = 2

\int_{1}^{\frac{3}{2}} ∣ 4 x - 4 ∣ d x = \frac{1}{2}

\int_{\frac{3}{2}}^{\frac{9}{4}} 4 x - \frac{4}{9} x^{2} \cdot 4 d x = \frac{7}{8}

= 4 + 2 + \frac{1}{2} + \frac{7}{8}

Vereinfache

= \frac{59}{8}

\int_{6}^{8} \frac{62}{( x - 6 ) ^{3}} d x

\frac{\partial}{\partial u} (u^{2} v)

\int 12 x^{3} (ln (4 x))^{2} d x

\int (3 x^{2} + 8) d x

y^{^{'}} = \frac{1 - 2 t}{y}