integral from 0 to 1 of 2sqrt(1-x^2)

Lösung

\int_{0}^{1} 2 1 - x^{2} d x

\frac{π}{2}

Dezimale

1.57079 \dots

Schritte zur Lösung

\int_{0}^{1} 2 1 - x^{2} d x

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{1} 1 - x^{2} d x

Trigonometrische Substitution anwenden

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= 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

= 2 \cdot \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

= 2 \cdot \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

= 2 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Vereinfache

2 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{π}{2}

= \frac{π}{2}

t a y l or (x^{2} + 1) e^{x - 1}, 1

\frac{d y}{d x} = \frac{( 4 sec ( y ) )}{( x + 3 ) ^{2}}

\frac{d}{d x} (\frac{( x ^{4} - 5 x ^{3} + x )}{x ^{2}})

\frac{d}{d x} (e^{4 x} tan (x^{5} - 4 x^{4}))

\int 9 sin^{2} (x co) s^{2} x d x