integral from 0 to 1 of ln(2)+y

Lösung

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

ln (2) + \frac{1}{2}

Dezimale

1.19314 \dots

Schritte zur Lösung

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

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

\int_{0}^{1} ln (2) d y = ln (2)

\int_{0}^{1} y d y = \frac{1}{2}

= ln (2) + \frac{1}{2}

\int_{a}^{b} (x - x^{2}) d x

\int_{- 2}^{3} y - (y^{2} - 6) d y

\int_{- 2}^{2} \frac{8}{x ^{2} + 4} d x

\int_{- 1}^{3} - x^{3} - 5 x + 1 d x

\int_{- 1}^{1} (2 x^{2} + 7) d x