integral from 0 to 7r of sqrt(49r^2-y^2)

Lösung

\int_{0}^{7 r} 49 r^{2} - y^{2} d y

\frac{49 π}{4} r^{2}

Schritte zur Lösung

\int_{0}^{7 r} 49 r^{2} - y^{2} d y

Trigonometrische Substitution anwenden

= \int_{0}^{\frac{π}{2}} 49 r^{2} cos^{2} (u) d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 49 r^{2} \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 49 r^{2} \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 49 r^{2} \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 u) d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 49 r^{2} \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= 49 r^{2} \frac{1}{2} (\frac{π}{2} + 0)

Vereinfache

49 r^{2} \frac{1}{2} (\frac{π}{2} + 0) : \frac{49 π}{4} r^{2}

= \frac{49 π}{4} r^{2}

\int ln (1 + r^{2}) r d r

\frac{\partial}{\partial x} (yz ln (x y))

y (x + y + 1) d x + (x + 2 y) d y = 0

x^{^{''}} - 2 x - 5 = 0, x (0) = 1, x (1) = 3

\int \frac{39}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x