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Q- the value of integration from (0 to -2)
(x(cube)+3x(sq.)+3+(x+1) cos(x+1) dx is
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
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