Question

The area of the region bounded by the parabola $x^2 = y$ and the line $y = x$ is

• A. $\frac{1}{2}$ sq. units
• B. $\frac{1}{6}$ sq. units
• C. $\frac{1}{3}$ sq. units
• D. $\frac{5}{6}$ sq. units

Hint:

There is an incredibly handy standard formula for the area trapped between a standard parabola $y^2 = 4ax$ and a line $y = mx$, or symmetrically $x^2 = 4by$ and $y = mx$. The enclosed area is always exactly given by $\frac{8 a^2}{3 m^3}$ or tracking the coefficients to yield $\frac{1}{6}$ for basic identity intersections. Memorizing this pattern saves huge integration layout steps!

Answer 1

The Correct Option is B

Step 1: Find where the curves meet.
The parabola is $y=x^2$ and the line is $y=x$. Set them equal to get the crossing points. $x^2=x$ gives $x(x-1)=0$, so $x=0$ and $x=1$.

Step 2: Decide which curve is on top.
Pick any value between 0 and 1, say $x=0.5$. The line gives $0.5$ and the parabola gives $0.25$. So the line stays above the parabola here.

Step 3: Set up the area integral.
Area is the integral of (top minus bottom) from 0 to 1. \[ A=\int_0^1 (x-x^2)\,dx \]

Step 4: Integrate and put in the limits.
\[ A=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6} \] So the area is $\frac{1}{6}$ square units.
\[ \boxed{\frac{1}{6}\text{ sq. units}} \]