Suppose we define integral using the following formula `int_(a)^(b)f(x)dx = (b-a)/(2) (f(a)+f(b))`, for more accurate result for `c in (a, b), F(c) = (c-a)/(2) (f(a)+f(c)) + (b-c)/(2)(f(b) + f(c))`.
When `c = (a+b)/(2)`, then `int_(a)^(b) f(x)dx = (b-a)/(4)(f(a) + f(b) + 2f(c))`.
`lim_(trarra)(int_(a)^(t)f(x) dx -((t-a))/(2)(f(t)+f(a)))/((t-a)^(3))=0 forall a` Then the degree of `f(x) ` can at most be

If f(a+b-x) =f(x) , then int_(a)^(b) x f(x) dx is equal to

int_(a+c)^(b+c) f(x) dx is equal to

Prove that int_(a)^(b)f(x)dx=(b-a)int_(0)^(1)f((b-a)x+a)dx

If int_(a)^(b)(f(x))/(f(x)+f(a+b-x))dx=10 , then

Evaluate each of the following integral: int_a^b(f(x))/(f(x)+f(a+b-x))dx

Let the definite integral be defined by the formula int_(a)^(b)f(x)dx=(b-a)/2(f(a)+f(b)) . For more accurate result, for c epsilon (a,b), we can use int_(a)^(b)f(x)dx=int_(a)^(c)f(x)dx+int_(c)^(b)f(x)dx=F(c) so that for c=(a+b)/2 we get int_(a)^(b)f(x)dx=(b-a)/4(f(a)+f(b)+2f(c)) . If f''(x)lt0 AA x epsilon (a,b) and c is a point such that altcltb , and (c,f(c)) is the point lying on the curve for which F(c) is maximum then f'(c) is equal to

If | int_(a)^(b) f(x)dx|= int_(a)^(b)|f(x)|dx,a ltb,"then " f(x)=0 has

If f(x) = a + bx + cx^2 where a, b, c in R then int_o ^1 f(x)dx

Prove that int_a^bf(x)dx=int_(a+c)^(b+c)f(x-c)dx , and when f(x) is odd function, int_(-a)^af(x)dx=0

If f(a+b-x)=f(x),\ then prove that \ int_a^b xf(x)dx=(a+b)/2int_a^bf(x)dx