Let `g(x) = cosx^(2), f(x) = sqrt(x)`, and `alpha, beta (alpha lt beta)` be the roots of the quadratic equation `18x^(2) -9pix + pi^(2) = 0`. Then the area (in sq. units) bounded by the curve `y = (gof) (x)` and the lines `x = alpha,x = beta` and `y = 0` isA. `1/2(sqrt(3) - sqrt(2))`B. `1/2(sqrt(3)-1)`C. `1/2(sqrt(3)-1)`D. `1/2(sqrt(3)+1)`

Let `g(x) = cosx^(2), f(x) = sqrt(x)`, and `alpha, beta (alpha lt beta

1.	Let `g(x) = cosx^(2), f(x) = sqrt(x)`, and `alpha, beta (alpha lt beta)` be the roots of the quadratic equation `18x^(2) -9pix + pi^(2) = 0`. Then the area (in sq. units) bounded by the curve `y = (gof) (x)` and the lines `x = alpha,x = beta` and `y = 0` isA. `1/2(sqrt(3) - sqrt(2))`B. `1/2(sqrt(3)-1)`C. `1/2(sqrt(3)-1)`D. `1/2(sqrt(3)+1)`
Answer» Correct Answer - C

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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Prove that:`I_n=int_0^oox^(2n+1)e^-x^2dx=(n !)/2,n in Ndot`
If f(x) is continuous and `int_(0)^(9)f(x)dx=4`, then the value of the integral `int_(0)^(3)x.f(x^(2))dx` isA. 2B. 18C. 16D. 4
`lim_(trarr0) int_(0)^(2pi)(|sin(x+t)-sinx|)/(|t|)dx` equalsA. 2B. 4C. 43469D. 1
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))`