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

Lösung

\int_{0}^{1} (2 - e^{x})^{2} d x

8 + \frac{e ^{2} - 1}{2} - 4 e

Dezimale

0.32140 \dots

Schritte zur Lösung

\int_{0}^{1} (2 - e^{x})^{2} d x

Wende U-Substitution an

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

Multipliziere aus

\frac{( 2 - u ) ^{2}}{u} : \frac{4}{u} - 4 + u

= \int_{1}^{e} \frac{4}{u} - 4 + 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

= \int_{1}^{e} \frac{4}{u} d u - \int_{1}^{e} 4 d u + \int_{1}^{e} u d u

\int_{1}^{e} \frac{4}{u} d u = 4

\int_{1}^{e} 4 d u = 4 e - 4

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

= 4 - (4 e - 4) + \frac{e ^{2} - 1}{2}

Vereinfache

= 8 + \frac{e ^{2} - 1}{2} - 4 e

\int_{3}^{1} 1

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