Question

Aplicind formula lui Leibniz Newton sa se calculeze integrala:

Answer 1

$\it \int_{-1}^1 x^4\ dx=\dfrac{\ x^5}{5} \mid_{-1}^1 = \dfrac{1^5}{5} - \dfrac{(-1)^5}{5} = \dfrac{1}{5} +\dfrac{1}{5} = \dfrac{2}{5} .$
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$\it \int_0 ^1 x(x-1)^2\ dx= \int_0 ^1 x(x^2-2x+1) \ dx= \int_0 ^1 (x^3-2x^2+x) \ dx =$

$\it \int_0 ^1 \ x^3\ dx - 2\int_0 ^1 x^2\ dx + \int_0 ^1 x\ dx = \dfrac{x^4}{4} \mid_0^1-\ 2\dfrac{x^3}{3} \mid_0^1 \ + \dfrac{x^2}{2} \mid_0^1 =$

$\it \dfrac{1}{4} -\dfrac{2}{3} +\dfrac{1}{2} =\dfrac{1}{12} \ .$