Question

Алгебра. Даю 20 баллов.

Answer 1

Ответ:

Объяснение:

$\displaystyle a)\;\;\; \int\limits^1_{-1} {x^4} \, dx =\frac{x^5}{5}\left|^1_{-1} =\frac{1^5}{5}-\frac{(-1)^5}{5}=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$

$\displaystyle b)\;\;\;\int\limits^3_{-1} {(1-2x+3x^2)} \, dx =(x-2*\frac{x^2}{2}+3*\frac{x^3}{3})|^3_{-1}=\\\\= (x-x^2+x^3)|^3_{-1}=(3-3^2+3^3)-(-1-(-1)^2+(-1)^3)=\\\\=3-9+27+1+1+1=24$

$\displaystyle c)\;\;\;\int\limits^{\frac{\pi }{6} }_{\frac{\pi }{12} } {sin\;2x} \, dx =\frac{1}{2}(-cos\;2x)|^{\frac{\pi }{6} } _{\frac{\pi }{12} }=-\frac{1}{2}(cos\frac{\pi }{3}-cos\frac{\pi }{6})=\\\\= -\frac{1}{2}\left(\frac{1}{2}-\frac{\sqrt{3} }{2}\right) =\frac{\sqrt{3}-1 }{4}$