integral from x to 1 of (x+y)

Lösung

\int_{x}^{1} (x + y) d y

x - x^{2} + \frac{1 - x ^{2}}{2}

Schritte zur Lösung

\int_{x}^{1} x + 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_{x}^{1} x d y + \int_{x}^{1} y d y

\int_{x}^{1} x d y = x - x^{2}

\int_{x}^{1} y d y = \frac{1 - x ^{2}}{2}

= x - x^{2} + \frac{1 - x ^{2}}{2}

\int_{1}^{2} (3 x - 5 x + 1) d x

\int_{0}^{s i n (x)} 2 t d t

\int_{0}^{π} (3 e^{x} + 8 sin (x)) d x

\int_{2}^{4} - \frac{6}{x ^{2}} d x

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