fläche f(x)=((25-x^2)^{1/2}), 4/3 x,5

Lösung

fläche

f (x) = ((25 - x^{2})^{\frac{1}{2}}), \frac{4}{3} x, 5

\frac{75}{8} - \frac{25}{2} arcsin (\frac{3}{5})

Dezimale

1.33123 \dots

Schritte zur Lösung

Um die Schnittpunkte zu finden, löse

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

((25 - x^{2})^{\frac{1}{2}}) = \frac{4}{3} x : x = 3

((25 - x^{2})^{\frac{1}{2}}) = 5 : x = 0

\frac{4}{3} x = 5 : x = \frac{15}{4}

= \int_{0}^{3} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{3}^{\frac{15}{4}} ∣ f_{3} (x) - f_{2} (x) ∣ d x

= \int_{0}^{3} ((25 - x^{2})^{\frac{1}{2}}) - 5 d x + \int_{3}^{\frac{15}{4}} 5 - \frac{4}{3} x d x

\int_{0}^{3} ((25 - x^{2})^{\frac{1}{2}}) - 5 d x = - \frac{25}{2} arcsin (\frac{3}{5}) + 9

\int_{3}^{\frac{15}{4}} 5 - \frac{4}{3} x d x = \frac{3}{8}

= - \frac{25}{2} arcsin (\frac{3}{5}) + 9 + \frac{3}{8}

Vereinfache

= \frac{75}{8} - \frac{25}{2} arcsin (\frac{3}{5})

\frac{\partial}{\partial x} (a e^{2 x})

\int \frac{1}{( a + b x ) ^{2}} d x

\int \frac{1}{16 x ^{4} - 1} d x

\int \frac{3 e ^{3 t}}{3 + e ^{3 t}} d t

\frac{d y}{d x} = sin (17 x)