Question

Нужна с алгеброй. Прямо сейчас .

Подробно и понятно

Answer 1

Объяснение:

1)

$\int\limits^{-2}_{-3} {3x^{-2}} \, dx =3*\int\limits^{-2}_{-3} {x^{-2}} \, dx=-3*x^{-1}\ |_{-3}^{-2} =-3*\frac{1}{x}\ |_{-3}^{-2}=-3*(\frac{1}{-2} -\frac{1}{-3})=\\=-3*(\frac{1}{3}-\frac{1}{2})=-3*\frac{2-3}{3*2}=-3*(-\frac{1}{6})=\frac{1}{2}=0,5.$

2)

$\int\limits^{32}_1 {\frac{1}{\sqrt[5]{x} } } \, dx=\int\limits^{32}_1 {x^{-\frac{1}{5}} } } \, dx=\frac{5}{4}*x^\frac{4}{5}\ |_1^{32}=\frac{5}{4}*(32^\frac{4}{5}-1^\frac{4}{5} )=\frac{5}{4}*(16-1)=\\=\frac{5*15}{4}=\frac{75}{4}=18,75.$

3)

$\int\limits^3_1 {(x^3+x)^2} \, dx =\int\limits^3_1 {(x^6+2x^4+x^2)} \, dx=\int\limits^3_1 {x^6} \, dx+2*\int\limits^3_1 {x^4} \, dx+\int\limits^3_1 {x^2} \, dx =\\=\frac{x^7}{7}\ |_1^3+2*\frac{x^5}{5}\ |_1^3+\frac{x^3}{3}\ |_1^3=\frac{3^7}{7}-\frac{1^7}{7} +2*(\frac{3^5}{5} -\frac{1^5}{5})+\frac{3^3}{3}-\frac{1^3}{3}=\\=\frac{2187}{7}-\frac{1}{7}+2*(\frac{243}{5}-\frac{1}{5} )+\frac{27}{3}-\frac{1}{3}=\frac{2186}{7}+\frac{484}{5} +\frac{26}{3}=\frac{2186*15+484*21+26*35}{7*5*3}=\\$

$=\frac{32790+10164+910}{105}=\frac{50864}{105} \approx484,42.$

Answer 2

1)3*x^(-2+1)/(-2+1)=-3/x=-3/(-2)-(-3/(-3))=3/2-1=1/2

2)∫(x^(1/5))dx=x^(1/5+1)/(1/5+1)=x^(6/5)/(6/5)=5x^(6/5)/6=

подстановка по х от 1 до 32

=5*32^(6/5)/6-5*1/6=5*64/6-5/6=(320-5)/6=315/6=52.5

3)∫(x^6+2x^4+x^2)dx=x^7/7+2*x^5/5+x^3/3=x^7/7+0.4x^5+x^3/3=

подстановка по х от 1 до 3

=3^7/7+0.4*3^5+3^3/3-1/7-0.4-1/3=312 3/7+97.2+9-1/7-0.4-1/3