integral from-3 to 1 of (2x+1)(3+x)(1-x)

Lösung

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

- \frac{32}{3}

Dezimale

- 10.66666 \dots

Schritte zur Lösung

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

Multipliziere mit der Konjugation von

2 x + 1 : \frac{4 x ^{2} - 1}{2 x - 1}

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

Bestimme unbestimmte (Singularitäts-)Punkte:

x = \frac{1}{2}

Wenn

b, a < b < c, f (b) =

unbestimmt

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

\int_{- 3}^{\frac{1}{2}} \frac{4 x ^{2} - 1}{2 x - 1} (3 + x) (1 - x) d x = - \frac{1127}{96}

\int_{\frac{1}{2}}^{1} \frac{4 x ^{2} - 1}{2 x - 1} (3 + x) (1 - x) d x = \frac{103}{96}

= - \frac{1127}{96} + \frac{103}{96}

Vereinfache

- \frac{1127}{96} + \frac{103}{96} : - \frac{32}{3}

= - \frac{32}{3}

\int_{- 1}^{2} - x^{2} + 4 d x

\int_{1}^{3} (5 x - x^{2}) d x

\int_{4}^{5} x^{2} d x

\int_{0}^{4} (4 - x) (x) d x

\int_{- 6}^{- 3} - x - 2 d x