This is a beautiful problem from ISI MStat 2016 Problem 1, PSB Sample, based on area bounded by the curves. We provide a detailed solution with the prerequisites mentioned explicitly.
In the diagram below, is a straight line that intersects the graph of a polynomial
of degree 2 at the points
and
The area of the shaded region is 36 square units. Obtain the expression for
.
Let as given
is of degree 2 .
Now from the figure we can see that intersect
at points
and
Hence we have and
, which gives ,
---(1) and
----(2)
Then ,
See from Fig-1 we can say that Area of the shaded region = (Area bounded by the curve P(x) and x-axis )- (Area of the triangle ABC) - (Area bounded by the curve P(x) , x=5 and x=L )
=
= -36
= -36
Again it is given that area of the shaded region is 36 square units.
So, -36 =36
=
. After integration we get ,
---(3)
Now we have three equations and three unknows
Solving this three equations by elimination and substitution we get ,
Therefore , the expression for is
.
This is a beautiful problem from ISI MStat 2016 Problem 1, PSB Sample, based on area bounded by the curves. We provide a detailed solution with the prerequisites mentioned explicitly.
In the diagram below, is a straight line that intersects the graph of a polynomial
of degree 2 at the points
and
The area of the shaded region is 36 square units. Obtain the expression for
.
Let as given
is of degree 2 .
Now from the figure we can see that intersect
at points
and
Hence we have and
, which gives ,
---(1) and
----(2)
Then ,
See from Fig-1 we can say that Area of the shaded region = (Area bounded by the curve P(x) and x-axis )- (Area of the triangle ABC) - (Area bounded by the curve P(x) , x=5 and x=L )
=
= -36
= -36
Again it is given that area of the shaded region is 36 square units.
So, -36 =36
=
. After integration we get ,
---(3)
Now we have three equations and three unknows
Solving this three equations by elimination and substitution we get ,
Therefore , the expression for is
.