Question

The area (in sq. units) of the region enclosed between the parabola y² = 2x and the line x + y = 4 is _______.

Answer 1

Correct Answer: 18

To find the area of the region enclosed between the parabola \(y^2=2x\) and the line \(x+y=4\), we first determine their points of intersection. Substitute \(x=4-y\) from the line equation into the parabola equation:
\(y^2=2(4-y)\)
\(y^2=8-2y\)
Rearranging gives \(y^2+2y-8=0\).
This quadratic equation can be solved using the quadratic formula \(y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), where \(a=1\), \(b=2\), and \(c=-8\).
Thus, \(y=\frac{-2\pm\sqrt{2^2-4\cdot1\cdot(-8)}}{2\cdot1}=\frac{-2\pm\sqrt{4+32}}{2}\).
\(y=\frac{-2\pm\sqrt{36}}{2}\)
\(y=\frac{-2\pm6}{2}\).
So, \(y=2\) or \(y=-4\).
Substituting back to find \(x\), for \(y=2\), \(x=4-2=2\), and for \(y=-4\), \(x=4-(-4)=8\).
The intersections are \((2,2)\) and \((8,-4)\).
Now, integrate to find the area between these curves from \(y=-4\) to \(y=2\).
The area \(A\) is given by:
\(A=\int_{-4}^{2}\left((4-y)-\frac{y^2}{2}\right)dy\).
Calculate the integral:
\(\int_{-4}^{2}(4-y)dy=\left[4y-\frac{y^2}{2}\right]_{-4}^{2}\).
Evaluating gives \(\left(8-2\right)-\left(-16-8\right)=6+24=30\).
For the parabola: \(\int_{-4}^{2}-\frac{y^2}{2}dy=-\frac{1}{2}\left[\frac{y^3}{3}\right]_{-4}^{2}\).
This evaluates to \(-\frac{1}{2}\left(\frac{8}{3}-\left(-\frac{64}{3}\right)\right)=-\frac{1}{2}\times\frac{72}{3}=-12\).
Thus, \(A=30-12=18\).
The area is 18 square units, which falls within the given range of 18,18.