integral from-1 to 4 of (3x-2)(x+3)

Lösung

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

\frac{175}{2}

Dezimale

87.5

Schritte zur Lösung

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

Multipliziere mit der Konjugation von

3 x - 2 : \frac{9 x ^{2} - 4}{3 x + 2}

= \int_{- 1}^{4} \frac{9 x ^{2} - 4}{3 x + 2} (x + 3) d x

Bestimme unbestimmte (Singularitäts-)Punkte:

x = - \frac{2}{3}

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_{- 1}^{- \frac{2}{3}} \frac{9 x ^{2} - 4}{3 x + 2} (x + 3) d x + \int_{- \frac{2}{3}}^{4} \frac{9 x ^{2} - 4}{3 x + 2} (x + 3) d x

\int_{- 1}^{- \frac{2}{3}} \frac{9 x ^{2} - 4}{3 x + 2} (x + 3) d x = - \frac{175}{54}

\int_{- \frac{2}{3}}^{4} \frac{9 x ^{2} - 4}{3 x + 2} (x + 3) d x = \frac{2450}{27}

= - \frac{175}{54} + \frac{2450}{27}

Vereinfache

- \frac{175}{54} + \frac{2450}{27} : \frac{175}{2}

= \frac{175}{2}

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

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

\int_{0}^{\frac{3 π}{2}} 5 sin (x) d x

\int_{0}^{2} (x + 1) (3 x - 4) d x

\int_{1}^{5} 2 x + 3 d x