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Magazine Chapter 5: Problem 49
Sketch the curve and find the total area between the curve and the given interval on the \(\bar{x}\) -axis. \(y=x^{2}-x ;[0,2]\)
Short Answer
Expert verified
The total area between the curve \(y = x^2 - x\) and the x-axis over the interval \([0,2]\) is \(\frac{2}{3}\).
Step by step solution
01
Understand the Problem
We are tasked with sketching the curve of the function \( y = x^2 - x \) and finding the total area between this curve and the x-axis over the interval \([0, 2]\).
02
Sketch the Curve
First, recognize that the given function \( y = x^2 - x \) is a quadratic function of the form \(y = ax^2 + bx + c\). Here, \(a = 1\), \(b = -1\), and \(c = 0\). This is a parabola opening upwards. Calculate the roots to find the x-intercepts by setting \(y = 0\):\[ x(x - 1) = 0 \]thus, \(x = 0\) and \(x = 1\). At these points, the curve intersects the x-axis. The vertex is at \(x = \frac{-b}{2a} = \frac{1}{2}\). Plot these points to sketch the curve between \(x = 0\) and \(x = 2\).
03
Set Up the Integral for Area Calculation
We will find the area under the curve \(y = x^2 - x\) from \(x = 0\) to \(x = 2\) using definite integration. Set up the integral:\[ \int_0^2 (x^2 - x) \, dx \]
04
Compute the Integral
Calculate the integral \( \int_0^2 (x^2 - x) \, dx \): Find the antiderivative: - The antiderivative of \(x^2\) is \(\frac{x^3}{3}\). - The antiderivative of \(x\) is \(\frac{x^2}{2}\).Thus, the antiderivative is \(\frac{x^3}{3} - \frac{x^2}{2}\). Evaluate this from 0 to 2: \[ \left[ \frac{2^3}{3} - \frac{2^2}{2} \right]_0^2 = \left(\frac{8}{3} - 2 \right) - \left(0 \right) = \frac{8}{3} - \frac{6}{3} = \frac{2}{3} \]
05
Interpret the Result
The result from the integration, \(\frac{2}{3}\), represents the total area between the curve and the x-axis over the interval \([0, 2]\). Since the curve dips below the x-axis between \(x = 0\) and \(x = 1\), this total incorporates both areas above and below the x-axis.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Function
A quadratic function is a type of polynomial function that is defined by an equation of the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The graph of a quadratic function is a parabola, which is a symmetrical curve.
- The parabola can open upwards or downwards depending on the sign of the coefficient \(a\). If \(a > 0\), the parabola opens upwards, like the function \(y = x^2 - x\) we are examining here.
- The quadratic function's vertex represents its maximum or minimum point. For an upward-opening parabola, the vertex is the minimum.
- Roots or x-intercepts of the quadratic function are determined by solving \(ax^2 + bx + c = 0\). These points indicate where the graph crosses the x-axis, as seen with \(x = 0\) and \(x = 1\) for this function.
Curve Sketching
Curve sketching is an essential technique in calculus for understanding the behavior of a function over a particular interval. To sketch the curve for a quadratic function like \(y = x^2 - x\), follow these steps:
- Determine key points: Start by identifying points like roots (where \(y = 0\)) and the vertex. For the function \(y = x^2 - x\), we identified the roots at \(x = 0\) and \(x = 1\), and the vertex at \(x = \frac{1}{2}\).
- Assess the direction of the parabola: As the coefficient \(a = 1\) is positive, this quadratic curves upwards.
- Analyze behavior at interval end points: Check how the curve behaves at given interval limits, \(x = 0\) and \(x = 2\), to conform with its opening direction and curvature.
Area Between Curve and Axis
Finding the area between a curve and the x-axis using definite integration is a foundational concept in calculus. You calculate this by evaluating the definite integral of the function over a specified interval. For the quadratic function \(y = x^2 - x\) over the interval \([0, 2]\):
- Set up the integral: Formulate the integral \(\int_0^2 (x^2 - x) \, dx\) to find the enclosed area between the curve and the x-axis.
- Compute the antiderivative: Here, the antiderivative of \(x^2\) is \(\frac{x^3}{3}\), and of \(x\) is \(\frac{x^2}{2}\). Thus, the integral becomes \(\left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_0^2\).
- This result, \(\frac{2}{3}\), represents the total area between the curve and the x-axis within the interval. It combines areas above and below the axis for accuracy.
