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Popular Cálculo >

integral de 0 a 1 de tsqrt(1+t^2+t^4)

Solução

\int_{0}^{1} t 1 + t^{2} + t^{4} d t

Solução

\frac{3 3 - 1}{8} + \frac{3}{16} ln (3 + 2) - \frac{3}{32} ln (3)

+1

Decimal

0.66845 \dots

Passos da solução

\int_{0}^{1} t 1 + t^{2} + t^{4} d t

Aplicar integração por substituição

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

Remover a constante:

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

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

Completar o cuadrado

1 + u + u^{2} : (u + \frac{1}{2})^{2} + \frac{3}{4}

= \frac{1}{2} \cdot \int_{0}^{1} (u + \frac{1}{2})^{2} + \frac{3}{4} d u

Aplicar integração por substituição

= \frac{1}{2} \cdot \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{4 v ^{2} + 3}{2} d v

Remover a constante:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{\frac{1}{2}}^{\frac{3}{2}} 4 v^{2} + 3 d v

Aplicar integração por substituição trigonométrica

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{3}{2} sec^{3} (w) d w

Remover a constante:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} sec^{3} (w) d w

Aplicar a redução de integrais

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} ([\frac{sec ^{2} ( w ) sin ( w )}{2}]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{3}} sec (w) d w)

\int_{\frac{π}{6}}^{\frac{π}{3}} sec (w) d w = ln (3 + 2) - \frac{1}{2} ln (3)

= \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} ([\frac{sec ^{2} ( w ) sin ( w )}{2}]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

Simplificar

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} ([\frac{sec ^{2} ( w ) sin ( w )}{2}]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3))) : \frac{3}{8} ([\frac{1}{2} sec (w) tan (w)]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

= \frac{3}{8} ([\frac{1}{2} sec (w) tan (w)]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

Calcular os limites:

3 - \frac{1}{3}

= \frac{3}{8} (3 - \frac{1}{3} + \frac{1}{2} (ln (3 + 2) - \frac{1}{2} ln (3)))

Simplificar

= \frac{3 3 - 1}{8} + \frac{3}{16} ln (3 + 2) - \frac{3}{32} ln (3)

Gráfico

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