integral from 0 to 2 of 2x-cos(4x)

Lösung

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

4 - \frac{1}{4} sin (8)

Dezimale

3.75266 \dots

Schritte zur Lösung

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

\int_{0}^{2} 2 x d x = 4

\int_{0}^{2} cos (4 x) d x = \frac{1}{4} sin (8)

= 4 - \frac{1}{4} sin (8)

\int_{0}^{9} 2 π x (x) d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1}{x ^{3}} d x

\int_{1}^{2} ln ((4 x)^{2}) d x

\int_{1}^{6} x^{4} ln (x) d x

\int_{0}^{\frac{y}{4}} e^{- y^{2}} d y