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 $P(x)$ .

ISI MStat 2016 Problem 1 figure

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 ,

ISI MStat 2016 Problem 1 graph — Fig-1

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 $P(x)= -x^2+6x+7$ .

Previous MStat Posts:

Related

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 $P(x)$ .

ISI MStat 2016 Problem 1 figure

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 ,

ISI MStat 2016 Problem 1 graph — Fig-1

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 $P(x)= -x^2+6x+7$ .

Previous MStat Posts:

Related

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