If `f and g` are continuous functions on `[0, a]` such that `f(x) = f(a - x) and g(x) + g(a - x)=2,` then show that `int_0^a f(x)g(x) dx = int_0^a f(x) dx`.A. `int_(0)^(a)f(x)dx`B. `int_(0)^(a)g(x)dx`C. 0D. `int_(0)^(a//2)f(x)dx`

If `f and g` are continuous functions on `[0, a]` such that `f(x) = f(

1.	If `f and g` are continuous functions on `[0, a]` such that `f(x) = f(a - x) and g(x) + g(a - x)=2,` then show that `int_0^a f(x)g(x) dx = int_0^a f(x) dx`.A. `int_(0)^(a)f(x)dx`B. `int_(0)^(a)g(x)dx`C. 0D. `int_(0)^(a//2)f(x)dx`
Answer» Correct Answer - A

`int_(0)^(4)(3sqrt(x+5))/(3sqrt(x+5)+3sqrt(9-x))dx=`A. 1B. 2C. 3D. 4
`int_(4)^(7)((11-x)^(2))/(x^(2)+(11-x)^(2))dx=`A. `(1)/(2)`B. 1C. `(3)/(2)`D. 2
`int_(-pi//2)^(pi//2)(sin^(4)x)/(sin^(4)x + cos^(4)x)dx=`A. 2B. `(pi)/(4)`C. `-2`D. `(pi)/(2)`
`int_(-2)^(2)(x^(3))/(4-x^(2))dx=`A. `(pi)/(2)-1`B. 0C. `1-(pi)/(2)`D. `log((2+e)/(2-e))`
Evaluate: `int_(1//4)^(1//2)1/(sqrt(x-x^2))dx`A. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)`
`int_(0)^(pi//2)sin^(3)x dx=`A. `(2)/(3)`B. `(3)/(2)`C. `-(2)/(3)`D. `-(3)/(2)`
`int_(a)^(b)(f(x))/(f(x)+f(a+b-x))dx=`A. `(a+b)/(2)`B. `(b-a)/(2)`C. `a+b`D. `b-a`
`int_(0)^(pi//2)(1)/(1+sqrt(tan x))dx=`A. `(pi)/(2)`B. `(pi)/(3)`C. `(pi)/(4)`D. `(pi)/(6)`
`int_(-1)^(1)(x^(2))/(1+x^(2))dx=`A. 1B. `pi`C. `2-pi`D. `tan^(-1)e`
If `int_(0)^(1)f(x)dx=1, int_(0)^(1)x f(x)dx=a` and `int_(0)^(1)x^(2)f(x)dx=a^(2)`, then : `int_(0)^(1)(a-x)^(2)f(x)dx=`A. `4a^(2)`B. 0C. `2a^(2)`D. `3a^(2)`