Question

интегралы,.............​

Answer 1

4.

$\int\limits^4_1 {\frac{2x^2+\frac{x}{2}+3 }{\sqrt{x}} } \, dx =\int\limits^4_1 {(\frac{2x^2+\frac{x}{2}+3 }{x^\frac{1}{2}}) } \, dx =\int\limits^4_1 {((2x^2+\frac{x}{2}+3)\cdot x^{-\frac{1}{2}}}) \, dx = \\ \\ =\int\limits^4_1 {(2x^\frac{3}{2}+\frac{x^\frac{1}{2}}{2}+3x^{-\frac{1}{2}}) \, dx =(2\cdot \frac{x^\frac{5}{2}}{\frac{5}{2}}+\frac{1}{2}\cdot \frac{x^\frac{3}{2}}{\frac{3}{2}}+3\cdot \frac{x^\frac{1}{2}}{\frac{1}{2}})|^4_1=(\frac{4}{5}\sqrt{x^5} +\frac{1}{3}\sqrt{x^3} +6\sqrt{x}})|^4_1=$

$=(\frac{4}{5}\sqrt{4^5} +\frac{1}{3}\sqrt{4^3} +6\sqrt{4} ) - (\frac{4}{5}\sqrt{1^5} +\frac{1}{3}\sqrt{1^3}+6\sqrt{1} )=\frac{4}{5}\cdot \sqrt{2^{10}}+\frac{1}{3}\sqrt{2^6}+12-\frac{4}{5}-\frac{1}{3}-6=\\ \\ =\frac{4}{5}\cdot 2^5+\frac{1}{3}\cdot 2^3+6-\frac{4}{5}-\frac{1}{3}=\frac{4}{5}\cdot 32+\frac{1}{3}\cdot 8+6-\frac{4}{5}-\frac{1}{3}=\frac{128}{5}-\frac{4}{5}+\frac{8}{3}-\frac{1}{3}+6=\\ \\ = \frac{124}{5}+\frac{7}{3}+6=\frac{372+35+90}{15}=\frac{497}{15}=33\frac{2}{15}$

5.

$\int\limits^7_2 {\frac{dx}{\sqrt{x+2}}} =\int\limits^7_2 {(x+2)^{-\frac{1}{2}}} \, dx = \int\limits^7_2 {(x+2)^{-\frac{1}{2}}} \, d(x+2) = \frac{(x+2)^\frac{1}{2}}{\frac{1}{2}}|^7_2=2\cdot \sqrt{x+2} |^7_2=\\ \\ = 2\cdot ( (\sqrt{7+2} ) - (\sqrt{2+2} ))=2\cdot (3-2)=2\cdot 1=2$

6.

$\int\limits^1_0 {x\cdot e^{-x}} \, dx =\left[\begin{array}{ccc}u=x; \ du=dx\\dv=e^{-x} \, dx; \ \ v=-e^{-x}\\\end{array}\right] =(x\cdot (e^{-x}))|^1_0-\int\limits^1_0 {(-e^{-x})} \, dx = \\ \\ = (-e^{-x}\cdot x)|^1_0+\int\limits^1_0 {e^{-x}} \, dx =( (-e^{-x}\cdot x) + (-e^{-x}))|^1_0=(e^{-x}\cdot (-x-1))|^1_0= \\ \\ = (e^{-1}\cdot (-1-1)}) - (e^{0}\cdot (0-1))=-\frac{2}{e}-(-1)=-\frac{2}{e}+1=1-\frac{2}{e}$