If the value of the definite integral `int_0^1^(2007)C_7x^(2000)dot(1-x)^7dx`is equal to `1/k ,w h e r ek in N ,`then the value of `k/(26)`is____

If the value of the definite integral `int_0^1^(2007)C_7x^(2000)dot(1-

1.	If the value of the definite integral `int_0^1^(2007)C_7x^(2000)dot(1-x)^7dx`is equal to `1/k ,w h e r ek in N ,`then the value of `k/(26)`is____
Answer» Correct Answer - 208 Let `I=int_(0)^(1).^(207)C_(7).ubrace(x^(200))_(II).ubrace((1-x)^(7))_(I)dx` `=.^(207)C_(7)[(1-x)^(7).(x^(201))/201\|_(0)^(1)+7/201int_(0)^(1)(1-x)^(6).x^(201)dx]` `=.^(207)C_(7) . 1/207 int_(0)^(1)(1-x)^(6) . x^(201)dx` Integrating by parts againsix times more we get `I=.^(207)C_(7).(7!)/(201.202.203.204.205.206.207)int_(0)^(1)x^(207)dx` `=((207)!)/(7!(200)!) . (7!)/(201.202......207) . 1/208` `=((207)!)/((207)!7!) . (7!)/208=1/208=1/k` or `k=208`

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`
Obtain the area enclosed by region bounded by the curves `y = x ln x` and ` y = 2x-2x^(2)`.A. `7//6`B. `7//24`C. `12//7`D. `7//12`
If `I_n=int_0^pi e^x(sinx)^n dx ,` then `(I_3)/(I_1)` is equal toA. `3//5`B. `1//5`C. `1`D. `2//5`
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))`