Определени интеграли

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Определени интеграли

Определени интеграли

20171206_230626.jpg (954.39 KiB) Прегледано 470 пъти

[aifC]

1.2 [tex]\int\limits_{4}^{12}\sqrt{x-3}dx; u=x-3; \int \sqrt{u}du = \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} = \frac{2}{3}(x-3)^{\frac{3}{2}}+C \Rightarrow 18 - \frac{2}{3} = \frac{52}{3};[/tex]

2.2[tex]\int\limits_{e}^{e^{2}}ln(ex)dx = \int (ln(x)+1)dx = \int ln(x)dx + \int 1dx = xlnx - \int 1dx + \int 1dx = xln(x)+C \Rightarrow
e^{2}ln(e^{2})-eln(e)=2e^{2}-e;[/tex]

3.2[tex]\int\limits_{0}^{4}\frac{x}{\sqrt{2x+1}}dx = \int \left(\frac{\sqrt{2x+1}}{2} - \frac{1}{2\sqrt{2x+1}}\right)= \frac{1}{2}\int \sqrt{2x+1}dx - \frac{1}{2}\int \frac{1}{\sqrt{2x+1}}dx = I_{1}-I_{2}; I_{1} = \int \sqrt{2x+1}dx; u=2x+1 \rightarrow dx = \frac{1}{2}du \Rightarrow \frac{1}{2} \int \sqrt{u}du =[/tex]
[tex]\frac{u^{\frac{3}{2}}}{3} \Rightarrow I_{1} = \frac{(2x+1)^{\frac{3}{2}}}{3}; I_{2} = \int \frac{1}{\sqrt{2x+1}}dx; u=2x+1, dx=\frac{1}{2}du = \frac{1}{2}\int \frac{1}{\sqrt{u}}du = \frac{\sqrt{2x+1}}{2} \Rightarrow I_{1}-I_{2} = \frac{(x-1)\sqrt{2x+1}}{3} +C \Rightarrow 3 - \left(- \frac{1}{3}\right) = \frac{10}{3};[/tex]

[aifC]

4.2[tex]y' = 3sin(x)+2 \Rightarrow \int 3sin(x)+2dx = \int 3sin(x) + \int 2dx = -3cos(x)+2x+C \Rightarrow y = -3cos(x)+2x+c_{1};[/tex]

5.3[tex]y'' = 3cos(x)+12x \Rightarrow y = y_{h}+y_{p}; ay''+by'+cy=0 \rightarrow e^{\gamma \cdot t} \rightarrow e^{ɣx}; e^{ɣx}(ɣ^{2}) = 0 \Rightarrow ɣ =
0; y_{h}=c_{1}+c_{2}x; y_{p} = y_{p1}+y_{p2}\Rightarrow y_{p1}= 2x^{3} + 0 \cdot x^{2} = 2x^{3}; y_{p2}= 0 \cdot sin(x)+(-3)cos(x) = -3cos(x); \Rightarrow y = c_{1}+c_{2}x+2x^{3}-3cps(x);[/tex]

[aifC]

10.2 [tex]x^{2}-4x+3 \Rightarrow x=3,1; \int\limits_{1}^{3}|x^{2}-4x+3-0|dx = -\frac{x^{3}}{3}+2x^{2}-3x+C \Rightarrow 0 - \left(-\frac{4}{3}\right)=\frac{4}{3};[/tex]

11.2 [tex]x^{2}-5x+6 \Rightarrow x=4,2; \int\limits_{2}^{4}|x^{2}-5x+6-(x-1)|dx = -\frac{x^{3}}{3}+3x^{2}-8x+C \Rightarrow -\frac{16}{3}- \left(-\frac{20}{3}\right) = \frac{4}{3};[/tex]

12.2 [tex](x-2)^{2}=x-2 \Rightarrow x=3,2 \int\limits_{2}^{3}|(x-2)^{2}-(x-2)|dx = -\frac{x^{3}}{3}+\frac{5x^{2}}{2}-6x+C \Rightarrow -\frac{9}{2} - \left(-\frac{14}{3}\right) = \frac{1}{6};[/tex]