`int[f(a x+b)]^nf^(prime)(a x+b)dx`

By substitution: Theorem: If int f(x)dx=phi(x) then int f(ax+b)dx=(1)/(a)phi(ax+b)dx

Prove that the equality int_(a)^(b) f(x) dx = int_(a)^(b) f(a + b - x) dx

int _(a) ^(b) f (x) dx = int _(a) ^(b) f (a + b - x) dx

If f(x) is monotonic differentiable function on [a,b], then int_(a)^(b)f(x)dx+int_(f(a))^(f(b))f^(-1)(x)dx=(a)bf(a)-af(b)(b) bf (b)-af(a)(c)f(a)+f(b)(d) cannot be found

int_0^(2a)f(x)dx is equal to 2int_0^af(x)dx b. 0 c. int_0^af(x)dx+int_0^af(2a-x)dx d. int_0^af(x)dx+int_0^(2a)f(2a-x)dx

If f(x)<0,\ x in (a , b) then at the point C(c ,\ f(c)) on y=f(x) for which F(c) is a maximum, f^(prime)(c) is given by a. f^(prime)(c)=(f(b)-f(a))/(b-1) b. \ f^(prime)(c)=(f(b)-f(a))/(a-b) c. f^(prime)(c)=(2(f(b)-f(a)))/(b-a) d. f^(prime)(c)=0

int_(a + c)^(b+c) f(x)dx=