integral from 0 to 1 of (-1+(pix)/2)

Lösung

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

- 1 + \frac{π}{4}

Dezimale

- 0.21460 \dots

Schritte zur Lösung

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

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

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

= - 1 + \frac{π}{4}

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

\int_{0}^{π} \frac{1}{x} d x

\int_{1}^{4} x^{3} ln (x) d x

\int_{0}^{1} x^{3} \cdot e^{- x} d x

\int_{1}^{3} x^{3} e^{- \frac{5}{3} x} d x