Question

The area (in square units) of the region \[A=\left\{ \left( x,y \right):\dfrac{{{y}^{2}}}{2}\le x\le y+4 \right\}\] is:
\[\left( a \right)\dfrac{53}{3}\]
\[\left( b \right)18\]
\[\left( c \right)30\]
\[\left( d \right)16\]

Answer 1

Hint: We are given two regions for which we have to find the area of their intersection. First, we will solve these equations to find the point of intersection on our first region is the parabola \[\dfrac{{{y}^{2}}}{2}=x\] and the other is x = y + 4. We will equate them to get the quadratic equation \[{{y}^{2}}-2y-8=0.\] We will solve and get the point of intersection as y = – 2 and y = 4. Then we will find the area under the graph by subtracting the area of the parabola from the area of the line on the limit – 2 to 4.

Complete step-by-step answer:
We are asked to find the area that lies between the two regions. The first region is \[x=\dfrac{{{y}^{2}}}{2}\] and the other region is x = y + 4. Clearly, the first region is a parabola symmetric to the x axis and the other one is a line.

We will find the point of their intersection for that we have to solve both the equations \[x=\dfrac{{{y}^{2}}}{2}\] and x = y + 4.
Now, equating them, we get,
\[\dfrac{{{y}^{2}}}{2}=y+4\]
Simplifying further we get,
\[\Rightarrow {{y}^{2}}=2y+8\]
\[\Rightarrow {{y}^{2}}-2y=8\]
We get a quadratic equation. We will factorize it using the middle term split.
\[\Rightarrow {{y}^{2}}-\left( 4-2 \right)y-8=0\]
\[\Rightarrow {{y}^{2}}-4y+2y-8=0\]
\[\Rightarrow y\left( y-4 \right)+2\left( y-4 \right)=0\]
\[\Rightarrow \left( y+2 \right)\left( y-4 \right)=0\]
So, we get y = – 2 and y = 4 where the two regions intersect.

Now, we can see that area under the graph of both the regions in the area that lies in the region y = – 2 and y = 4. To find our required area, we have to subtract the area under the line by area under the parabola. This means,
Required Area = Area under the line – Area under the Parabola
\[\Rightarrow \text{Required Area}=\int\limits_{-2}^{4}{\left( y+4 \right)dy}-\int\limits_{-2}^{4}{\left( \dfrac{{{y}^{2}}}{2} \right)dy}\]
Applying the limits, we get,
\[\Rightarrow \text{Required Area}=\left[ \dfrac{{{4}^{2}}}{2}+4\times 4 \right]-\left[ \dfrac{{{\left( -2 \right)}^{2}}}{2}+4\times 2 \right]-\left[ \dfrac{{{4}^{3}}}{6}-\dfrac{{{\left( -2 \right)}^{3}}}{6} \right]\]
Simplifying further, we get,
\[\Rightarrow \text{Required Area}=\left[ \left( 8+16 \right)-\left( 2-8 \right) \right]-\left[ \dfrac{32}{3}+\dfrac{4}{3} \right]\]
\[\Rightarrow \text{Required Area}=30-12\]
\[\Rightarrow \text{Required Area}=18sq.units\]

So, the correct answer is “Option B”.

Note: We will find the point of intersection, we will get a quadratic equation which we will solve by the middle term split. We split our middle term in such a way that the product of these will give us the number the same as constant term product with the coefficient of \[{{x}^{2}}.\] We will always subtract the area of the region that lies below from the region that lies above.