Question

Решите пожалуйста 3-4 варианты

Answer 1

1.

1)

$\int\limits^1_0 {\frac{dx}{1+x^2} } =\int\limits^1_0 {\frac{dx}{1+x^2} } = arctg \, x |^1_0= arctg \, 1 - arctg \, 0 = \frac{\pi}{4}-0=\frac{\pi}{4}$

2)

$\int\limits^1_{-1} {5(y^2+1)} \, dy=5\int\limits^1_{-1} {(y^2+1)} \, dy=5\cdot (\frac{y^3}{3}+y)|^1_{-1}=5\cdot ((\frac{1^3}{3}+1)-(\frac{(-1)^3}{3}-1))=\\ \\ =5\cdot (\frac{1}{3}+1+\frac{1}{3}+1)=5\cdot \frac{8}{3}=\frac{40}{3}$

3)

$\int\limits^3_2 {(2x-1)^2} \, dx =\int\limits^3_2 {(4x^2-4x+1)} \, dx=(4\cdot \frac{x^3}{3}-4\cdpt \frac{x^2}{2}+x)|^3_2=(\frac{4x^3}{3}-2x^2+x)|^3_2= \\ \\ = (\frac{4}{3}\cdot 3^3 -2\cdot 3^2 +3 ) - (\frac{4}{3}\cdot 2^3-2\cdot 2^2 +2 )=(36-18+3)-(\frac{32}{3}-8+2)=21-\frac{14}{3}=\frac{49}{3}$

2.

1) $y=9-x^2, \ \ y=0, \ \ x=-1, \ x=2$

$\int\limits^2_{-1} {((9-x^2)-0)} \, dx=\int\limits^2_{-1} {(9-x^2)} \, dx =(9x-\frac{x^3}{3})|^2_{-1}= \\ \\ = (9\cdot 2-\frac{2^3}{3})-(9\cdot (-1)-\frac{(-1)^3}{3})=(18-\frac{8}{3})-(-9+\frac{1}{3})=27-3=24$

2)

$y=x, \ \ y=2-x^2\\\\ x=2-x^2\\ \\ x^2+x-2=0 \\ \\ x_{1.2}=\frac{-1\pm \sqrt{1^2-4\cdot 1 \cdot (-2)}}{2\cdot 1}=\frac{-1\pm 3}{2}; \\\\ x_1 = \frac{-1+3}{2}=1; \ \ \ \ x_2=\frac{-1-3}{2}=-2$

$\int\limits^1_{-2} {(2-x^2-x)} \, dx =(2x-\frac{x^3}{3}-\frac{x^2}{2})|^1_{-2}=(2\cdot 1 -\frac{1^3}{3}-\frac{1^2}{2})-(2\cdot (-2)-\frac{(-2)^3}{3}-\frac{(-2)^2}{2})=(2-\frac{1}{3}-\frac{1}{2})-(-4+\frac{8}{3}-2)=\frac{7}{6}+\frac{10}{3}=\frac{27}{6}=\frac{9}{2}$

1.

1)

$\int\limits^\frac{\sqrt{3}}{2}_{-1} {\frac{dx}{\sqrt{1-x^2}} = \arcsin {x} |^\frac{\sqrt{3}}{2}_{-1}=\arcsin{\frac{\sqrt{3}}{2}}-\arcsin{(-1)}=\frac{\pi}{3}-(-\frac{\pi}{2})=\frac{5\pi}{6}$

2)

$\int\limits^0_{-2} {2\cdot(x^3-x)} \, dx= 2 \int\limits^0_{-2} {(x^3-x)} \, dx =2\cdot (\frac{x^4}{4}-\frac{x^2}{2})|^0_{-2}=2\cdot (0-(\frac{(-2)^4}{4}-\frac{(-2)^2}{2}))=\\ \\ =2\cdot (-4+2)=-4$

3)

$\int\limits^2_{-1} {(x^2-1)^3} \, dx =\int\limits^2_{-1} {(x^6-3x^4+3x^2-1)} \, dx =(\frac{x^7}{7}-3\cdot \frac{x^5}{5}+3\cdot \frac{x^3}{3}-x)|^2_{-1}= \\ \\ =(\frac{2^7}{7}-\frac{3\cdot 2^5}{5}+2^3-2)-(-\frac{1}{7}+\frac{3}{5}-1+1)=(\frac{128}{7}-\frac{96}{5}+8-2)+\frac{1}{7}-\frac{3}{5}=\\ \\ =\frac{129}{7}-\frac{99}{5}+6=\frac{645-693+210}{35}=\frac{162}{35}$

2.

1)

$y=3x^2+1, \ \ y=0, \ \ x=-1; \ \ x=2 \\\\\\\int\limits^2_{-1} {(3x^2+1-0)} \, dx =(x^3+x)|^2_{-1}=(2^3+2)-((-1)^3+(-1))=10+2=12$

2)

$y=x^2-4x+5, \ \ x=x+5 \\ \\ x^2-4x+5=x+5 \\ \\ x^2 -5x=0 \\ \\ x\cdot (x-5)=0 \\ \\ x_1=0; \ \ \ \ x_2-5=0 \\ \\ x_1=0; \ \ \ \ x_2=5$

$\int\limits^5_0 {((x+5)-(x^2-4x+5))} \, dx =\int\limits^5_0 {(5x-x^2)} \, dx =(5\cdot \frac{x^2}{2}-\frac{x^3}{3})|^5_0=\frac{125}{2}-\frac{125}{3}-0=\\ \\ =\frac{375-250}{6}=\frac{125}{6}$