What is the area included in the first quadrant between the curves y = x and y = x 3 ?
Finding Area Between Curves in First Quadrant
The problem asks us to find the area bounded by the curves \(y = x\) and \(y = x^3\) in the first quadrant. To solve this, we need to identify the region, find the points of intersection, determine which function is above the other in the relevant interval, and then use integration to calculate the area.
Step 1: Find the Points of Intersection
To find where the curves \(y = x\) and \(y = x^3\) intersect, we set the equations equal to each other:
\[x = x^3\]Rearranging the equation gives:
\[x^3 - x = 0\]Factor out \(x\):
\[x(x^2 - 1) = 0\]Factor the difference of squares:
\[x(x-1)(x+1) = 0\]The solutions are the values of \(x\) where the curves intersect. These are:
- \(x = 0\)
- \(x - 1 = 0 \implies x = 1\)
- \(x + 1 = 0 \implies x = -1\)
The points of intersection occur at \(x = -1, 0, 1\). Since the question specifies the area in the first quadrant, we are interested in the interval \(x \ge 0\). The relevant intersection points in the first quadrant are at \(x = 0\) and \(x = 1\).
Step 2: Determine the Upper and Lower Curves
We need to determine which function, \(y = x\) or \(y = x^3\), has a greater value over the interval \(0 \le x \le 1\). Let's pick a test value within this interval, for example, \(x = 0.5\).
- For \(y = x\), the value is \(0.5\).
- For \(y = x^3\), the value is \((0.5)^3 = 0.125\).
Since \(0.5 > 0.125\), the curve \(y = x\) is above \(y = x^3\) for \(0 < x < 1\). Therefore, \(f(x) = x\) is the upper function and \(g(x) = x^3\) is the lower function in the interval of interest.
Step 3: Set up the Definite Integral for Area
The area between two curves \(y = f(x)\) and \(y = g(x)\) from \(x = a\) to \(x = b\), where \(f(x) \ge g(x)\) over \([a, b]\), is given by the definite integral:
\[\text{Area} = \int_{a}^{b} (f(x) - g(x)) \,dx\]In our case, \(a = 0\), \(b = 1\), \(f(x) = x\), and \(g(x) = x^3\). So the integral for the area is:
\[\text{Area} = \int_{0}^{1} (x - x^3) \,dx\]Step 4: Evaluate the Integral
Now, we evaluate the definite integral:
\[\text{Area} = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1}\]Evaluate the antiderivative at the upper limit (\(x=1\)) and subtract the value at the lower limit (\(x=0\)):
\[\text{Area} = \left( \frac{(1)^2}{2} - \frac{(1)^4}{4} \right) - \left( \frac{(0)^2}{2} - \frac{(0)^4}{4} \right)\] \[\text{Area} = \left( \frac{1}{2} - \frac{1}{4} \right) - (0 - 0)\] \[\text{Area} = \frac{1}{2} - \frac{1}{4}\]To subtract the fractions, find a common denominator, which is 4:
\[\text{Area} = \frac{2}{4} - \frac{1}{4}\] \[\text{Area} = \frac{2 - 1}{4}\] \[\text{Area} = \frac{1}{4}\]The area included in the first quadrant between the curves \(y = x\) and \(y = x^3\) is \(\frac{1}{4}\) square unit.
Let's check this against the given options:
| Option | Value |
|---|---|
| 1 | \(\frac{1}{8}\) square unit |
| 2 | \(\frac{1}{4}\) square unit |
| 3 | \(\frac{1}{2}\) square unit |
| 4 | 1 square unit |
Our calculated area matches Option 2.
Revision Table: Area Between Curves
| Concept | Description | Application in this Problem |
|---|---|---|
| Finding Intersection Points | Set the equations of the curves equal to each other and solve for the variable (usually \(x\)). | \(x = x^3 \implies x = 0, 1, -1\). Relevant for first quadrant: \(x=0, 1\). |
| Identifying the Region | Based on the intersection points and any constraints (like "first quadrant"), determine the interval(s) of integration. | First quadrant means \(x \ge 0, y \ge 0\). The interval is \(x \in [0, 1]\). |
| Determining Upper vs. Lower Curve | Pick a test point within the interval and evaluate both functions to see which has a larger value. | At \(x=0.5\), \(y=x\) is 0.5, \(y=x^3\) is 0.125. \(y=x\) is the upper curve. |
| Setting up Integral for Area | Integrate the difference between the upper and lower functions over the determined interval. \(\int_{a}^{b} (f_{upper}(x) - f_{lower}(x)) \,dx\) | \(\int_{0}^{1} (x - x^3) \,dx\) |
| Evaluating Definite Integral | Find the antiderivative of the integrand and evaluate it at the upper and lower limits, then subtract. \([F(x)]_{a}^{b} = F(b) - F(a)\) | \(\left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{4}\) |
Additional Information: Area Calculation using Integration
Calculating the area between curves is a fundamental application of definite integrals in calculus. The general idea is to slice the region into infinitely thin vertical or horizontal strips and sum their areas. For vertical strips, as used in this problem, the width of each strip is \(dx\), and the height is the difference between the \(y\)-values of the upper and lower curves at that \(x\). This leads to the integral formula \(\int_{a}^{b} (f_{upper}(x) - f_{lower}(x)) \,dx\).
It's crucial to correctly identify the interval of integration, usually determined by the points where the curves intersect or by given boundaries. It's also vital to know which function is the upper curve and which is the lower curve within that specific interval. If the curves cross within the desired region, you might need to split the area calculation into multiple integrals over sub-intervals where the upper/lower relationship remains constant.
In this problem, the curves \(y=x\) and \(y=x^3\) intersect at \(x=0\) and \(x=1\) in the first quadrant, neatly defining a single region for integration. The curve \(y=x\) is a straight line through the origin, and \(y=x^3\) is a cubic function that is below \(y=x\) for \(0 < x < 1\) but crosses it at \(x=1\) and becomes greater for \(x > 1\).
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