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kanika kaul Grade: 12
        

 


Q- the value of  integration from (0 to -2)


(x(cube)+3x(sq.)+3+(x+1) cos(x+1) dx is

7 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

I=x^3+3x^2+3+(x+1)cos(x+1)


 ={x^3+1+3x(x+1)} +2-3x+(x+1)cos(x+1)          adding and subtracting -3x


 ={(x+1)^3)}+ 5 -3(x+1) +(x+1)cos(x+1)         


now put x+1=t


           so dx=dt


now becomes


              I=(t^3 -3t+tcost+5)lim +1to-1                   (t^3,t,tcost all are odd functions so integral will be 0)


              I=5(1+1)


                =10

7 years ago
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