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

Lösung

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

\frac{62}{3}

Dezimale

20.66666 \dots

Schritte zur Lösung

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

Multipliziere aus

(2 x - 1)^{2} : 4 x^{2} - 4 x + 1

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

Wende die Summenregel an:

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

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

\int_{1}^{3} 4 x^{2} d x = \frac{104}{3}

\int_{1}^{3} 4 x d x = 16

\int_{1}^{3} 1 d x = 2

= \frac{104}{3} - 16 + 2

Vereinfache

\frac{104}{3} - 16 + 2 : \frac{62}{3}

= \frac{62}{3}

\int_{- 2}^{2} 19 - ∣ x ∣^{3} d x

\int_{0}^{π \frac{1}{2}} e^{- x^{2}} d x

\int_{- 4}^{0} x^{3} 256 - x^{4} d x

\int_{0}^{x} (4 - t^{2}) e^{t^{2}} d t

\int_{x}^{1} y d y