There are two curves in a graph. One is y = x2 and the other is y = x. Find the area enclosed between these curves.
1 / 6 unit
Area Between Curves Explained
Finding the area enclosed between two curves is a common problem in calculus. This involves identifying the points where the curves intersect, determining which curve is above the other in the relevant interval, and then integrating the difference of the functions over that interval.
Understanding the Curves
We are given two equations representing curves:
- First curve: \(y = x\) (This is a straight line passing through the origin with a slope of 1).
- Second curve: \(y = x^2\) (This is a parabola opening upwards with its vertex at the origin).
Finding Intersection Points
To find the area enclosed between these curves, we first need to determine where they intersect. We do this by setting their y-values equal to each other:
\[ x^2 = x \]Now, rearrange the equation to solve for \(x\):
\[ x^2 - x = 0 \]Factor out \(x\):
\[ x(x - 1) = 0 \]This gives us two possible values for \(x\):
- \(x = 0\)
- \(x - 1 = 0 \implies x = 1\)
These are the x-coordinates of the points where the two curves intersect. So, the area we are interested in is enclosed between \(x=0\) and \(x=1\).
Identifying the Upper and Lower Curves
Between the intersection points \(x=0\) and \(x=1\), we need to determine which curve is "above" the other. We can pick a test value within this interval, for example, \(x = 0.5\).
- For \(y = x\), at \(x = 0.5\), \(y = 0.5\).
- For \(y = x^2\), at \(x = 0.5\), \(y = (0.5)^2 = 0.25\).
Since \(0.5 > 0.25\), the line \(y = x\) is above the parabola \(y = x^2\) in the interval \([0, 1]\). Therefore, \(y_{\text{upper}} = x\) and \(y_{\text{lower}} = x^2\).
Calculating the Enclosed Area
The area \(A\) enclosed between two curves \(y_{\text{upper}}(x)\) and \(y_{\text{lower}}(x)\) from \(x=a\) to \(x=b\) is given by the definite integral:
\[ A = \int_{a}^{b} (y_{\text{upper}}(x) - y_{\text{lower}}(x)) dx \]In our case, \(a = 0\), \(b = 1\), \(y_{\text{upper}}(x) = x\), and \(y_{\text{lower}}(x) = x^2\).
Substitute these into the integral formula:
\[ A = \int_{0}^{1} (x - x^2) dx \]Now, we integrate term by term:
\[ A = \left[ \frac{x^{1+1}}{1+1} - \frac{x^{2+1}}{2+1} \right]_{0}^{1} \] \[ A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} \]Next, apply the limits of integration (Fundamental Theorem of Calculus):
\[ A = \left( \frac{(1)^2}{2} - \frac{(1)^3}{3} \right) - \left( \frac{(0)^2}{2} - \frac{(0)^3}{3} \right) \] \[ A = \left( \frac{1}{2} - \frac{1}{3} \right) - (0 - 0) \] \[ A = \frac{1}{2} - \frac{1}{3} \]To subtract these fractions, find a common denominator, which is 6:
\[ A = \frac{3}{6} - \frac{2}{6} \] \[ A = \frac{3 - 2}{6} \] \[ A = \frac{1}{6} \]Therefore, the area enclosed between the curves \(y = x\) and \(y = x^2\) is \(\frac{1}{6}\) square unit.
The final answer is \(\frac{1}{6}\) unit.
Important Questions from Application of Integrals
What is the area of one of the loops between the curve y = c sin x and x-axis?
- The area enclosed between the curves $y = -x^2 + 4x$ and $y = x^2 - 2x$ is
The equation of the normal at the point (1, 1) on the curve 2y + x2 = 3 is
The area cut off the parabola 4y = 3x2 by the straight line 2y = 3x + 12 is
The area enclosed between the curves y2 = x and y = |x| is:
