integral from x^2 to x^4 of sin(sqrt(t))

Lösung

\int_{x^{2}}^{x^{4}} sin (t) d t

2 (- x^{2} cos (x^{2}) + sin (x^{2}) + x cos (x) - sin (x))

Schritte zur Lösung

\int_{x^{2}}^{x^{4}} sin (t) d t

Wende U-Substitution an

= \int_{x^{2}}^{x^{4}} sin (u) \cdot 2 u d u

Entferne die Konstante:

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

= 2 \cdot \int_{x^{2}}^{x^{4}} sin (u) u d u

Wende die partielle Integration an

= 2 [- u cos (u) - \int - cos (u) d u]_{x^{2}}^{x^{4}}

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

= 2 [- u cos (u) - (- sin (u))]_{x^{2}}^{x^{4}}

Vereinfache

2 [- u cos (u) - (- sin (u))]_{x^{2}}^{x^{4}} : 2 [- u cos (u) + sin (u)]_{x}^{x^{2}}

= 2 [- u cos (u) + sin (u)]_{x}^{x^{2}}

Berechne die Grenzen:

- x^{2} cos (x^{2}) + sin (x^{2}) + x cos (x) - sin (x)

= 2 (- x^{2} cos (x^{2}) + sin (x^{2}) + x cos (x) - sin (x))

\int_{0}^{1 - y^{2}} x y^{2} d x

\int_{1}^{4} 2 x - 3 d x

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

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