Class 12 Math Calculus Area

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)andy=f(c)(wherec∈(a,b))is minimum whenc=2a+b $.

$Proof :A=a∫ c (f(c)−f(x))dx+c∫ b (f(c))dx$

$=f(c)(c−a)−a∫ c (f(x))dx+a∫ b (f(x))dx−f(c)(b−c)$

$⇒A=[2c−(a+b)]f(c)+c∫ b (f(x))dx−a∫ c (f(x))dx$

Differentiating w.r.t. c, we get

$dcdA =[2c−(a+b)]f_{′}(c)+2f(c)+0−f(c)−(f(c)−0)$

For maxima and minima , $dcdA =0$

$⇒f_{′}(c)[2c−(a+b)]=0(asf_{′}(c)=0)$

Hence, $c=2a+b $

$Also forc<2a+b ,dcdA <0and forc>2a+b ,dcdA >0$

Hence, A is minimum when $c=2a+b $.

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

$Proof :A=a∫ c (f(c)−f(x))dx+c∫ b (f(c))dx$

$=f(c)(c−a)−a∫ c (f(x))dx+a∫ b (f(x))dx−f(c)(b−c)$

$⇒A=[2c−(a+b)]f(c)+c∫ b (f(x))dx−a∫ c (f(x))dx$

Differentiating w.r.t. c, we get

$dcdA =[2c−(a+b)]f_{′}(c)+2f(c)+0−f(c)−(f(c)−0)$

For maxima and minima , $dcdA =0$

$⇒f_{′}(c)[2c−(a+b)]=0(asf_{′}(c)=0)$

Hence, $c=2a+b $

$Also forc<2a+b ,dcdA <0and forc>2a+b ,dcdA >0$

Hence, A is minimum when $c=2a+b $.

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

Similar topics

relations and functions

integrals

trigonometric functions

inverse trigonometric functions

application of derivatives

relations and functions

integrals

trigonometric functions

inverse trigonometric functions

application of derivatives

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