Помогите решить алгебру!
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Ответил moboqe
0
1) ![1)intlimits^3_1 {-x^2+2x+3} , dx =-1/3x^3|limits^3_1+x^2|limits^3_1+3x|limits^3_1=-1/3(27-1)+\
+(9-1)+3(3-1)= -26/3+8+6=14-26/3=(42-26)/3=\
=16/3](https://tex.z-dn.net/?f=+1%29intlimits%5E3_1+%7B-x%5E2%2B2x%2B3%7D+%2C+dx+%3D-1%2F3x%5E3%7Climits%5E3_1%2Bx%5E2%7Climits%5E3_1%2B3x%7Climits%5E3_1%3D-1%2F3%2827-1%29%2B%5C%0A%2B%289-1%29%2B3%283-1%29%3D+-26%2F3%2B8%2B6%3D14-26%2F3%3D%2842-26%29%2F3%3D%5C%0A%3D16%2F3%0A "1)intlimits^3_1 {-x^2+2x+3} , dx =-1/3x^3|limits^3_1+x^2|limits^3_1+3x|limits^3_1=-1/3(27-1)+\
+(9-1)+3(3-1)= -26/3+8+6=14-26/3=(42-26)/3=\
=16/3")
<br/>
2)![intlimits^{1/3}_{-1} {(1-2x-3x^2)} , dx =x|limits^{1/3}_{-1}-x^2|limits^{1/3}_{-1}-x^3|limits^{1/3}_{-1}=(1/3+1)-\
-(1/9-1)-(1/27+1)=32/27](https://tex.z-dn.net/?f=+intlimits%5E%7B1%2F3%7D_%7B-1%7D+%7B%281-2x-3x%5E2%29%7D+%2C+dx+%3Dx%7Climits%5E%7B1%2F3%7D_%7B-1%7D-x%5E2%7Climits%5E%7B1%2F3%7D_%7B-1%7D-x%5E3%7Climits%5E%7B1%2F3%7D_%7B-1%7D%3D%281%2F3%2B1%29-%5C%0A-%281%2F9-1%29-%281%2F27%2B1%29%3D32%2F27 "intlimits^{1/3}_{-1} {(1-2x-3x^2)} , dx =x|limits^{1/3}_{-1}-x^2|limits^{1/3}_{-1}-x^3|limits^{1/3}_{-1}=(1/3+1)-\
-(1/9-1)-(1/27+1)=32/27")
<br/>
3)![intlimits^2_{-1} {(2-x^2+x)} , dx = 2x|limits^2_{-1}-1/3x^3|limits^2_{-1}+1/2x^2|limits^2_{-1}=\
=2(2+1)-1/3(8+1)+1/2(4-1)=4,5](https://tex.z-dn.net/?f=+intlimits%5E2_%7B-1%7D+%7B%282-x%5E2%2Bx%29%7D+%2C+dx+%3D+2x%7Climits%5E2_%7B-1%7D-1%2F3x%5E3%7Climits%5E2_%7B-1%7D%2B1%2F2x%5E2%7Climits%5E2_%7B-1%7D%3D%5C%0A%3D2%282%2B1%29-1%2F3%288%2B1%29%2B1%2F2%284-1%29%3D4%2C5 "intlimits^2_{-1} {(2-x^2+x)} , dx = 2x|limits^2_{-1}-1/3x^3|limits^2_{-1}+1/2x^2|limits^2_{-1}=\
=2(2+1)-1/3(8+1)+1/2(4-1)=4,5")
2)
3)
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Ответил moboqe
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