The area enclosed between the straight line y = x and the parabola y = x² in the x – y plane is____________

Area Calculation: Understanding the Problem

The problem asks us to find the area enclosed between two common mathematical functions in the x – y plane: a straight line and a parabola. Specifically, we need to find the area between the line $y=x$ and the parabola $y=x^2$.

To find the area enclosed between two curves, we typically follow these steps:

1. Find the points of intersection of the two curves. These points will define the limits of integration.
2. Determine which function is the "upper" curve and which is the "lower" curve within the interval defined by the intersection points.
3. Set up the definite integral using the formula $\text{Area}={\int }_a^b(y_{\text{upper}}-y_{\text{lower}})dx$.
4. Evaluate the definite integral to find the enclosed area.

Intersection Points: Finding the Limits of Integration

First, let's find where the line $y=x$ and the parabola $y=x^2$ intersect. We do this by setting their y-values equal to each other:

$x=x^2$

To solve for $x$, we can rearrange the equation:

$x^2-x=0$

Factor out $x$:

$x(x-1)=0$

This gives us two possible values for $x$:

- $x=0$
- $x-1=0\Rightarrow x=1$

These two x-values, $0$ and $1$, are our limits of integration.

Curve Analysis: Identifying Upper and Lower Functions

Now, we need to determine which function is above the other between $x=0$ and $x=1$. We can pick a test point within this interval, for example, $x=0.5$.

- For the line $y=x$: If $x=0.5$, then $y=0.5$.
- For the parabola $y=x^2$: If $x=0.5$, then $y=(0.5{)}^2=0.25$.

Since $0.5>0.25$, the line $y=x$ is the upper curve and the parabola $y=x^2$ is the lower curve in the interval $[0,1]$.

Area Integral: Setting Up the Calculation

With the limits of integration and the upper and lower curves identified, we can set up the definite integral to calculate the enclosed area. The formula for the area between two curves is:

$\text{Area}={\int }_a^b(y_{\text{upper}}-y_{\text{lower}})dx$

In our case, $a=0$, $b=1$, $y_{\text{upper}}=x$, and $y_{\text{lower}}=x^2$. So, the integral becomes:

$\text{Area}={\int }_0^1(x-x^2)dx$

Integral Evaluation: Solving for the Enclosed Area

Now, we evaluate the definite integral:

$\text{Area}={\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1$

First, substitute the upper limit $x=1$:

$\left(\frac{1^2}{2}-\frac{1^3}{3}\right)=\left(\frac{1}{2}-\frac{1}{3}\right)$

Next, substitute the lower limit $x=0$:

$\left(\frac{0^2}{2}-\frac{0^3}{3}\right)=(0-0)=0$

Subtract the value at the lower limit from the value at the upper limit:

$\text{Area}=\left(\frac{1}{2}-\frac{1}{3}\right)-0$

To subtract the fractions, find a common denominator, which is 6:

$\text{Area}=\left(\frac{3}{6}-\frac{2}{6}\right)$

$\text{Area}=\frac{3-2}{6}$

$\text{Area}=\frac{1}{6}$

Thus, the area enclosed between the straight line $y=x$ and the parabola $y=x^2$ is $\frac{1}{6}$ square units.