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# ISI MStat 2016 Problem 1 | Area bounded by the curves | PSB Sample

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.

## Problem- ISI MStat 2016 Problem 1

In the diagram below, $$L(x)$$ is a straight line that intersects the graph of a polynomial $$P(x)$$ of degree 2 at the points $$A=(-1,0)$$ and $$B=(5,12) .$$ The area of the shaded region is 36 square units. Obtain the expression for $$P(x)$$.

## Prerequisites

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

## Solution

Let $$P(x)=ax^2 +bx + c$$ as given $$P(x)$$ is of degree 2 .

Now from the figure we can see that $$L(x)$$ intersect $$P(x)$$ at points $$A=(-1,0)$$ and $$B=(5,12) .$$

Hence we have $$P(-1)=0$$ and $$P(5)=12$$ , which gives ,

$$a-b+c=0$$ ---(1) and $$25a+5b +c =12$$ ----(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 )

= $$\int^{L}_{-1} P(x)\,dx - \frac{1}{2} \times (5+1) \times 12 -\int^{5}_{L} P(x)\,dx$$

=$$\int^{L}_{-1} P(x)\,dx - \int^{5}_{L} P(x)\,dx$$ -36

=$$\int^{5}_{-1} P(x)\,dx$$ -36

Again it is given that area of the shaded region is 36 square units.

So, $$\int^{5}_{-1} P(x)\,dx$$ -36 =36 $$\Rightarrow$$ $$\int^{5}_{-1} P(x)\,dx$$ =$$2 \times 36$$

$$\int^{5}_{-1} (ax^2+bx+c) \,dx = 2 \times 36$$ . After integration we get ,

$$7a + 2b +c =12$$ ---(3)

Now we have three equations and three unknows

$$a-b+c=0$$

$$25a+5b +c =12$$

$$7a + 2b +c =12$$

Solving this three equations by elimination and substitution we get ,

$$a=-1 , b=6 , c=7$$

Therefore , the expression for $$P(x)$$ is $$P(x)= -x^2+6x+7$$ .

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