integral from 0 to 1 of a+bx^2

Lösung

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

a + \frac{b}{3}

Schritte zur Lösung

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

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

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

= a + \frac{b}{3}

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

\int_{0}^{\frac{π}{2}} \frac{1}{1 + sin ( x )} d x

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

\int_{0}^{2} x ln (x + 1) d x

\int_{- \infty}^{0} x 5^{- x} d x