Задача 57843 помогите, пожалуйста...
Условие
Решение
[red]2)[/red][m]\int_2^{1.5}{\frac{(2x-3)^\frac{1}{3}-(2x-3)^\frac{1}{4}}{(2x-3)^\frac{1}{5}}}dx=\int_2^{1.5}{(\frac{(2x-3)^\frac{1}{3}}{(2x-3)^\frac{1}{5}}}-\frac{(2x-3)^\frac{1}{4}}{(2x-3)^\frac{1}{5}})dx=\int_2^{1.5}{((2x-3)^\frac{2}{15}-(2x-3)^\frac{1}{20})}dx=\int_2^{1.5}{(2x-3)^\frac{2}{15}}dx-\int_2^{1.5}{(2x-3)^\frac{1}{20}}=[/m]
=[m]\frac{1}{2}\int_2^{1.5}{(2x-3)^\frac{2}{15}}d(2x-3)-\frac{1}{2}\int_2^{1.5}{(2x-3)^\frac{1}{20}}d(2x-3)=\frac{1}{2}*\frac{15(2x-3)^\frac{17}{15}}{17}-\frac{1}{2}*\frac{20(2x-3)^\frac{21}{20}}{21}=\frac{15(2x-3)^\frac{17}{15}}{34}-\frac{10(2x-3)^\frac{21}{20}}{21}=[/m]
=[m](\frac{15(2*1.5-3)^\frac{17}{15}}{34}-\frac{21(2*1.5-3)^\frac{21}{20}}{21})-(\frac{15(2*2-3)^\frac{17}{15}}{34}-\frac{10(2*2-3)^\frac{21}{20}}{21})=\frac{10}{21}-\frac{15}{34}=\frac{25}{714}[/m]
[red]3)[/red][m]\int_0^2{\frac{dx}{\sqrt{x^2+5}}}=\int_0^2{\frac{dx}{\sqrt{x^2+\sqrt{5}^2}}}=ln(x+\sqrt{x^2+\sqrt{5}^2})|_0^2=ln(x+\sqrt{x^2+5})|_0^2=ln(2+\sqrt{2^2+5})-ln(0+\sqrt{0^2+5})=ln(5)-ln(\sqrt{5})=[/m]
=[m]ln(5)-ln(5^\frac{1}{2})=ln(5)-\frac{1}{2}ln(5)=\frac{1}{2}ln(5)[/m]
[red]4)[/red][m]\int_{-1}^0{\frac{dx}{1-3x}}=-\frac{1}{3}\int_{-1}^0{\frac{d(1-3x)}{1-3x}}=-\frac{1}{3}*ln(1-3x)|_{-1}^0=-\frac{1}{3}(ln(1-3*0)-ln(1-3*(-1)))=0+\frac{1}{3}ln(4)=\frac{1}{3}ln(4)=\frac{1}{3}ln(2^2)=\frac{2}{3}ln(2)[/m]
[red]5)[/red] первый рисунок во вложении!
S=[m]\int_0^4(x-(x^2-3x))dx=\int_0^4(x-x^2+3x)dx=\int_0^4xdx-\int_0^4x^2dx+\int_0^43xdx=\int_0^4xdx-\int_0^4x^2dx+3\int_0^4xdx=\frac{x^2}{2}-\frac{x^3}{3}+\frac{3x^2}{2}|_0^4=[/m]
=[m](\frac{4^2}{2}-\frac{4^3}{3}+\frac{3*4^2}{2})-(\frac{0^2}{2}-\frac{0^3}{3}+\frac{3*0^2}{2})=(32-\frac{64}{3})-0=\frac{32}{3}[/m]
6)второй рисунок во вложении!
[m]\int_1^{1.5}(-x^2+2-(x^2-3x+2))dx=\int_1^{1.5}(-x^2+2-x^2+3x-2)dx=\int_1^{1.5}(-2x^2+3x)dx=\int_1^{1.5}-2x^2dx+\int_1^{1.5}3xdx=[/m]
=[m]-2\int_1^{1.5}x^2dx+3\int_1^{1.5}xdx=\frac{-2x^3}{3}+\frac{3x^2}{2}|_1^{1.5}=(\frac{-2*1.5^3}{3}+\frac{3*1.5^2}{2})-(\frac{-2*1^3}{3}+\frac{3*1^2}{2})=\frac{9}{8}-\frac{5}{6}=\frac{7}{24}[/m]
