integral from 0 to 1 of xcos(2x^2+7)

Lösung

\int_{0}^{1} x cos (2 x^{2} + 7) d x

\frac{1}{4} (sin (9) - sin (7))

Dezimale

- 0.06121 \dots

Schritte zur Lösung

\int_{0}^{1} x cos (2 x^{2} + 7) d x

Wende U-Substitution an

= \int_{7}^{9} \frac{cos ( u )}{4} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{7}^{9} cos (u) d u

Nutze das gemeinsame Integral :

\int cos (u) d u = sin (u)

= \frac{1}{4} [sin (u)]_{7}^{9}

Berechne die Grenzen:

sin (9) - sin (7)

= \frac{1}{4} (sin (9) - sin (7))

\frac{\partial}{\partial x} (x y^{2} + y^{2})

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s l o p e (- 4, - 4), (2, 5)

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