`int_(a)^(b)(f(x)dx)/(f(x)+f(a+b-x))=(1)/(2)(b-a)`

int_(0)^(2a)(f(x)dx)/(f(x)+f(2a-x))=a

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

f(x)>0AAx in R and is bounded. If lim_(n->oo)[int_0^a(f(x)dx)/(f(x)+f(a-x))+aint_a^(2a)(f(x)dx)/(f(x)+f(3a-x)) +a^2int_(2a)^(3a)(f(x)dx)/(f(x)+f(5a-x))+...+a^(n-1)int_((n-1)a)^(n a)(f(x)dx)/(f(x)+f[(2n-1)a-x]]] =7//5 (where a<1), then a is equal to

If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at x=a, x=b, y=f(x) and y=f(c)("where "c in (a,b))" is minimum when "c=(a+b)/(2) . "Proof : " A=int_(a)^(c)(f(c)-f(x))dx+int_(c)^(b)(f(c))dx =f(c)(c-a)-int_(a)^(c)(f(x))dx+int_(a)^(b)(f(x))dx-f(c)(b-c) rArr" "A=[2c-(a+b)]f(c)+int_(c)^(b)(f(x))dx-int_(a)^(c)(f(x))dx Differentiating w.r.t. c, we get (dA)/(dc)=[2c-(a+b)]f'(c)+2f(c)+0-f(c)-(f(c)-0) For maxima and minima , (dA)/(dc)=0 rArr" "f'(c)[2c-(a+b)]=0(as f'(c)ne 0) Hence, c=(a+b)/(2) "Also for "clt(a+b)/(2),(dA)/(dc)lt0" and for "cgt(a+b)/(2),(dA)/(dc)gt0 Hence, A is minimum when c=(a+b)/(2) . If the area bounded by f(x)=(x^(3))/(3)-x^(2)+a and the straight lines x=0, x=2, and the x-axis is minimum, then the value of a is

If f(x) is monotonic differentiable function on [a , b] , then int_a^bf(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

If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at x=a, x=b, y=f(x) and y=f(c)("where "c in (a,b))" is minimum when "c=(a+b)/(2) . "Proof : " A=int_(a)^(c)(f(c)-f(x))dx+int_(c)^(b)(f(c))dx =f(c)(c-a)-int_(a)^(c) (f(x))dx+int_(a)^(b)(f(x))dx-f(c)(b-c) rArr" "A=[2c-(a+b)]f(c)+int_(c)^(b)(f(x))dx-int_(a)^(c)(f(x))dx Differentiating w.r.t. c, we get (dA)/(dc)=[2c-(a+b)]f'(c)+2f(c)+0-f(c)-(f(c)-0) For maxima and minima , (dA)/(dc)=0 rArr" "f'(c)[2c-(a+b)]=0(as f'(c)ne 0) Hence, c=(a+b)/(2) "Also for "clt(a+b)/(2),(dA)/(dc)lt0" and for "cgt(a+b)/(2),(dA)/(dc)gt0 Hence, A is minimum when c=(a+b)/(2) . If the area enclosed by f(x)= sin x + cos x, y=a between two consecutive points of extremum is minimum, then the value of a is

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

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

Prove that the value of the integral, int_(0)^(2a)(f(x))/(f(x)+f(2a-x))dx is equal to a.

The value of int_(0)^(2a) (f(x))/(f(x)+f(2a-x))dx is equal to -