integral de 1 a 2 de sqrt(2x-x^2)

Solución

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

\frac{π}{4}

Pasos de solución

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

Completar el cuadrado

2 x - x^{2} : - (x - 1)^{2} + 1

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

Aplicar integración por sustitución

: \int_{0}^{1} - u^{2} + 1 d u

= \int_{0}^{1} - u^{2} + 1 d u

Aplicar integración por sustitución trigonométrica

: \int_{0}^{\frac{π}{2}} cos^{2} (v) d v

= \int_{0}^{\frac{π}{2}} cos^{2} (v) d v

Re-escribir usando identidades trigonométricas

= \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 v )}{2} d v

Sacar la constante:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 v) d v

Aplicar la regla de la suma:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d v + \int_{0}^{\frac{π}{2}} cos (2 v) d v)

\int_{0}^{\frac{π}{2}} 1 d v = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 v) d v = 0

= \frac{1}{2} (\frac{π}{2} + 0)

Simplificar

\frac{1}{2} (\frac{π}{2} + 0) : \frac{π}{4}

= \frac{π}{4}

\int_{0}^{1} 4 e^{- 2 x} d x

\int_{0}^{1} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

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\int_{0}^{t} e^{- a x} d x

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