Let `f(x)` and `phi(x)` are two continuous function on `R` satisfying `phi(x)=int_(a)^(x)f(t)dt, a!=0` and another continuous function `g(x)` satisfying `g(x+alpha)+g(x)=0AA x epsilonR, alpha gt0`, and `int_(b)^(2k)g(t)dt` is independent of `b` If `f(x)` is an odd function, thenA. `phi(x)` is also an odd functionB. `phi(x)` is an even functionC. `phi(x)` is neither an even nor an odd functionD. for `phi(x)` to be an even function, it must satisfy `int_(0)^(a)f(x)dx=0`

Let `f(x)` and `phi(x)` are two continuous function on `R` satisfying

1.	Let `f(x)` and `phi(x)` are two continuous function on `R` satisfying `phi(x)=int_(a)^(x)f(t)dt, a!=0` and another continuous function `g(x)` satisfying `g(x+alpha)+g(x)=0AA x epsilonR, alpha gt0`, and `int_(b)^(2k)g(t)dt` is independent of `b` If `f(x)` is an odd function, thenA. `phi(x)` is also an odd functionB. `phi(x)` is an even functionC. `phi(x)` is neither an even nor an odd functionD. for `phi(x)` to be an even function, it must satisfy `int_(0)^(a)f(x)dx=0`
Answer» Correct Answer - B `f(x)` is an odd function. Thus, `f(x)=-f(-x)` `phi(-x)=int_(0)^(-x)f(t)dt` Put `t=-y` `:. phi(-x)=int_(-a)^(x)f(-t)(-dt)` `=int_(-a)^(x)f(t)dt=int_(-a)^(a)f(t)dt+int_(a)^(x)f(t)dt` `=0+int_(a)^(x)f(t)dt=phi(x)`.

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))`