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

Lösung

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

\frac{4 2 - 2}{3}

Dezimale

1.21895 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{2}^{1} - u d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{1}^{2} - u d u

Entferne die Konstante:

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

= - (- \int_{1}^{2} u d u)

Wende die Potenzregel an

= - (- [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{2})

Vereinfache

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{2}

Berechne die Grenzen:

\frac{4 2}{3} - \frac{2}{3}

= \frac{4 2}{3} - \frac{2}{3}

Vereinfache

= \frac{4 2 - 2}{3}

\int_{- 1}^{3} - x^{3} + 3 x^{2} + 1 d x

\int_{x}^{x^{2}} \frac{1}{t ^{3} + 2} d t

\int_{0}^{8} 4 d x

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

\int_{0}^{x} \frac{( 36 - x ^{2} )}{144} d x