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=overset(c)underset(a)int(f(c)-f(x))dx+overset(b)underset(c)int(f(c))dx`
`=f(c)(c-a)-overset(c)underset(a)int(f(x))dx+overset(b)underset(a)int(f(x))dx-f(c)(b-c)`
`rArr" "A=[2c-(a+b)]f(c)+overset(b)underset(c)int(f(x))dx-overset(c)underset(a)int(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 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 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

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) . The value of the parameter a for which the area of the figure bounded by the abscissa axis, the graph of the function y=x^(3)+3x^(2)+x+a , and the straight lines, which are parallel to the axis of ordinates and cut the abscissa axis at the point of extremum of the function, which is the least, is

int(f'(x))/(f(x))dx=log f(x)+c

Let int(dx)/(sqrt(x^(2)+1)-x)=f(x)+C such that f(0)=0 and C is the constant of integration, then the value of f(1) is

If f is a continuous function on the interval [a,b] and there exists some c in(a,b) then prove that int_(a)^(b)f(x)dx=f(c)(b-a)

if int(dx)/(f(x))=log(f(x))^(2)+C then

if int(dx)/(f(x))=log(f(x))^(2)+C then