If `fa n dg`are continuous function on `[0,a]`satisfying `f(x)=f(a-x)a n dg(x)(a-x)=2,`then show that`int_0^af(x)g(x)dx=int_0^af(x)dxdot`

If `fa n dg`are continuous function on `[0,a]`satisfying `f(x)=f(a-x)a

1.	If `fa n dg`are continuous function on `[0,a]`satisfying `f(x)=f(a-x)a n dg(x)(a-x)=2,`then show that`int_0^af(x)g(x)dx=int_0^af(x)dxdot`
Answer» Correct Answer - NA `int_(0)^(a)f(x)g(x)dx` `=int_(0)^(a)f(a-x)g(a-x)dx` `=int_(0)^(a)f(x).{2-g(x)}dx` `=2int_(0)^(a)f(x)dx-int_(0)^(a)f(x)g(x)dx` or `2int_(0)^(a)f(x)f(x)dx=2int_(0)^(a)f(x)dx` or `int_(0)^(a)f(x)g(x)dx=int_(0)^(a)f(x)dx`

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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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))`