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, \(L(x)\) 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 \(P(x)\).
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 \( P(x)= -x^2+6x+7 \) .
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, \(L(x)\) 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 \(P(x)\).
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 \( P(x)= -x^2+6x+7 \) .
