Let `f"[0,oo)rarr R` be a continuous and stricity increasing function such that `f^(3) (x) =int_(0)^(x) tf^(2)(t) dt, x ge0`. The area enclosed by `y = f(x)` , the x-axis and the ordinate at `x = 3` is ` "_______"`A. `3/2`B. `5/2`C. `7/2`D. `1/2`

Let `f"[0,oo)rarr R` be a continuous and stricity increasing function

1.	Let `f"[0,oo)rarr R` be a continuous and stricity increasing function such that `f^(3) (x) =int_(0)^(x) tf^(2)(t) dt, x ge0`. The area enclosed by `y = f(x)` , the x-axis and the ordinate at `x = 3` is ` "_______"`A. `3/2`B. `5/2`C. `7/2`D. `1/2`
Answer» Correct Answer - A

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.
The area bounded by the curves `y=cosx` and `y=sinx` between the ordinates `x=0` and `x=(3pi)/2` isA. `4sqrt(2) + 2`B. `4sqrt(2) +1`C. `4sqrt(2) + 1`D. `4sqrt(2) - 2`
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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))`