If `int_0^3 (3ax^2+2bx+c)dx=int_1^3 (3ax^2+2bx+c)dx` where `a,b,c` are constants then `a+b+c=`A. `a+b+c=3`B. `a+b+c=1`C. `a+b+c=0`D. `a+b+c=2`

If `int_0^3 (3ax^2+2bx+c)dx=int_1^3 (3ax^2+2bx+c)dx` where `a,b,c` are

1.	If `int_0^3 (3ax^2+2bx+c)dx=int_1^3 (3ax^2+2bx+c)dx` where `a,b,c` are constants then `a+b+c=`A. `a+b+c=3`B. `a+b+c=1`C. `a+b+c=0`D. `a+b+c=2`
Answer» Correct Answer - C `int_(0)^(3)(3ax^(2)+2bx+c)dx=int_(1)^(3)(3ax^(2)+2bx+c)dx` `rArr" "int_(0)^(1)(3ax^(2)+2bx+c)dx+int_(1)^(3)(3ax^(2)+2bx+c)dx` `" "=int_(1)^(3)(3ax^(2)+2bx+c)` `rArr" "int_(0)^(1)(3ax^(2)+2bx+c)dx=0` `rArr" "[(3ax^(3))/(3)+(2bx^(2))/(2)+cx]_(0)^(1)=0` `rArr" "a+b+c=0`

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)` Making the substitution `"tan"1/2x=t`, we have `I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.
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Prove that:`I_n=int_0^oox^(2n+1)e^-x^2dx=(n !)/2,n in Ndot`
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Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.
`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`
`lim_(nrarroo) (((n+1)(n+2)"........"3n)/(n^(2n)))^(1//n)` is equal to :A. `(27)/(e^(2))`B. `(9)/(e^(2))`C. `3 log3-2`D. `(18)/(e^(4))`