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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 .

- Area bounded by the curve
- Polynomials of degree 2
- Area of a triangle

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 .

- Area bounded by the curve
- Polynomials of degree 2
- Area of a triangle

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 .

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