integral from 0 to 1 of 1-x^2+(x^4)/2

Lösung

\int_{0}^{1} 1 - x^{2} + \frac{x ^{4}}{2} d x

\frac{23}{30}

Dezimale

0.76666 \dots

Schritte zur Lösung

\int_{0}^{1} 1 - x^{2} + \frac{x ^{4}}{2} 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_{0}^{1} 1 d x - \int_{0}^{1} x^{2} d x + \int_{0}^{1} \frac{x ^{4}}{2} d x

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

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} \frac{x ^{4}}{2} d x = \frac{1}{10}

= 1 - \frac{1}{3} + \frac{1}{10}

Vereinfache

1 - \frac{1}{3} + \frac{1}{10} : \frac{23}{30}

= \frac{23}{30}

\int_{0}^{\frac{π}{6}} cos^{8} (3 A) d A

\int_{9}^{16} \frac{1}{1 + x} d x

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

\int_{0}^{h} (k x + v) d x

\int_{0}^{1} 2 x + x^{2} d x