Calculating Enclosed Area Between Curves
The problem asks for the area enclosed between the straight line $y = x$ and the parabola $y = x^2$. To find this area, we first need to determine the points where these two curves intersect.
1. Find Intersection Points
Set the equations equal to each other:
$ x = x^2 $
Rearrange the equation:
$ x^2 - x = 0 $
Factor out $x$:
$ x(x - 1) = 0 $
This gives two intersection points at $x = 0$ and $x = 1$.
2. Determine the Upper and Lower Functions
Within the interval $[0, 1]$, we need to know which function has a greater value. Let's test a point within the interval, for example, $x = 0.5$.
- For $y = x$: $y = 0.5$
- For $y = x^2$: $y = (0.5)^2 = 0.25$
Since $0.5 > 0.25$, the line $y = x$ is above the parabola $y = x^2$ in the interval $(0, 1)$.
3. Set Up and Evaluate the Integral
The area $A$ enclosed between the curves is calculated by integrating the difference between the upper function ($y = x$) and the lower function ($y = x^2$) over the interval $[0, 1]$.
The integral is:
$ A = \int_{0}^{1} (\text{upper function} - \text{lower function}) \, dx $
$ A = \int_{0}^{1} (x - x^2) \, dx $
Now, evaluate the definite integral:
$ A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} $
Substitute the limits of integration:
$ A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) $
$ A = \left( \frac{1}{2} - \frac{1}{3} \right) - (0 - 0) $
$ A = \frac{3}{6} - \frac{2}{6} $
$ A = \frac{1}{6} $
The area enclosed between the line $y = x$ and the parabola $y = x^2$ is $\frac{1}{6}$.
Important Questions from Application Of Definite Integral (Area)
The equation of a closed curve in two-dimensional polar coordinates is given by $r = \frac{2}{\sqrt{\pi}}(1 - \sin \theta)$. The area enclosed by the curve is ______ (answer in integer).
The area bounded by the curves, $y = \sqrt{x}$, and $y = 8x^2$ is _______________(rounded off to 3 decimal places)
Two straight lines pass through the origin $(x_0, y_0) = (0,0)$. One of them passes through the point $(x_1, y_1) = (1,3)$ and the other passes through the point $(x_2, y_2) = (1,2)$.
What is the area enclosed between the straight lines in the interval $[0, 1]$ on the x-axis?
Consider the equation for a curve, $y = f(x) = x^2 + x$.
The area enclosed by the curve, the x -axis ($y = 0$ line); the vertical lines passing through $x = 1$ and $x = 2$ is _________ (rounded off to 2 decimal places)- The area of the region (rounded off to one decimal place) enclosed between the curves $y = x$ and $y = 3\sqrt{x}$ and between the lines $x = 0$ and $x = 1$ is ________ units.
