Integrales Regla Barrow

Calcula las integrales de las siguientes funciones entre x=2 y x=4 aplicando la regla de Barrow:

- f(x)=x^5
- f(x)=3x^2-5x+6
- f(x)=\frac{-1}{2x}
- f(x)=2^x
- f(x)=\sqrt{x+2}

SOLUCIÓN

\int_2^4 x^5 \: dx = \left[ \frac{x^6}{6} \right]_2^4 = \frac{4^6}{6} - \frac{2^6}{6} = 672

\int_2^4 (3x^2-5x+6) \: dx= \left[ \frac{3x^3}{3}-\frac{5x^2}{2}+6x \right]_2^4 =
\left( 4^3-\frac{5 \cdot 4^2}{2}+6\cdot4 \right) - \left(2^3-\frac{5\cdot 2^2}{2}+6 \cdot 2 \right) =
(64-40+24) - (8-10+12) = 48-10=38

\int_2^4 \frac{-1}{2x} \: dx=\left[ -\frac{1}{2} Ln(x) \right]_2^4=
=\left(  -\frac{1}{2} Ln(4) \right) - \left(  -\frac{1}{2} Ln(2) \right) =\frac{Ln(2)-Ln(4)}{2}

\int_2^4 2^x \: dx=\left[ \frac{2^x}{Ln(2)} \right]_2^4=
=\left( \frac{2^4}{Ln(2)} \right) - \left( \frac{2^2}{Ln(2)} \right) =\frac{12}{Ln(2)}

\int_2^4 \sqrt{x+2} \: dx=\int_2^4 (x+2)^\frac{1}{2} \: dx=\left[ \frac{(x+2)^\frac{3}{2}}{\frac{3}{2}}  \right]_2^4=
=\left[ \frac{\sqrt{(x+2)^3}}{\frac{3}{2}} \right]_2^4=\left[ \frac{2 \cdot \sqrt{(x+2)^3}}{3} \right]_2^4
=\left( \frac{2 \cdot \sqrt{(4+2)^3}}{3} \right) - \left( \frac{2 \cdot \sqrt{(2+2)^3}}{3} \right) =
=\left( \frac{2 \cdot \sqrt{6^3}}{3} \right) - \left( \frac{2 \cdot \sqrt{4^3}}{3} \right) =
= \frac{2 \cdot \sqrt{6^3}}{3}  - \frac{2 \cdot \sqrt{4^3}}{3}  =
= \frac{12 \cdot \sqrt{6}}{3}  - \frac{16}{3}  =\frac{12\sqrt{6}-16}{3}