integral de 0 a 4 de sqrt(16-x^2)

Solución

\int_{0}^{4} 16 - x^{2} d x

4 π

Notación decimal

12.56637 \dots

Pasos de solución

\int_{0}^{4} 16 - x^{2} d x

Aplicar integración por sustitución trigonométrica

: \int_{0}^{\frac{π}{2}} 16 cos^{2} (u) d u

= \int_{0}^{\frac{π}{2}} 16 cos^{2} (u) d u

Sacar la constante:

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

= 16 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

Re-escribir usando identidades trigonométricas

= 16 \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

Sacar la constante:

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

= 16 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (2 u) d u

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

= 16 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

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

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

= 16 \cdot \frac{1}{2} (\frac{π}{2} + 0)

Simplificar

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

= 4 π

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