Question

Integrala de la 5 la 6 din 1 supra x-4 totul la patrat ????

Answer 1

$\int\limits^6_5 { \frac{1}{(x-4)^2} } \, dx=- \frac{1}{x-4}$$I\limits^6_5$=-[1/(6-4)-1/(5-4)]=1/2
 derivata lui $\frac{1}{x-4}$ este $- \frac{1}{(x-4)^2}$, deci invers integrala este $\frac{1}{x-4}$

Answer 2

[tex]\it I = \int\limits^6_5{\dfrac{1}{(x-4)^2}}\ dx [/tex]

Notam x - 4 = t ⇒ dx = dt  și vom avea:

[tex]\it \int{\dfrac{1}{(x-4)^2}}\ dx = \int{\dfrac{1}{t^2}}\ dt = \int{t^{-2}} \ dt = \dfrac{t^{-2+1}}{-2+1} +C = \\ \\ =\dfrac{t^{-1}}{-1} +C = -\dfrac{1}{x-4} +C[/tex]


$\it I = \int\limits^6_5{\dfrac{1}{(x-4)^2}}\ dx = -\dfrac{1}{x-4} |^6_5 =-\dfrac{1}{6-4} + \dfrac{1}{5-4} =-\dfrac{1}{2} +1 =\dfrac{1}{2}$