Chapter 5: Problem 46
Sketch the region described and find its area. The region below the curve \(y=x-x^{2}\) and above the \(x\) axis.
Short Answer
Expert verified
The area of the region is \( \frac{1}{6} \) square units.
Step by step solution
01
Identify the Curve and Region
The given curve is described by the equation \( y = x - x^2 \). This is a downward-facing parabola because the coefficient of \( x^2 \) is negative. We're tasked with finding the area of the region under this curve and above the \( x \)-axis.
02
Find the Intersections with the X-Axis
To find where the curve intersects the \( x \)-axis, we solve the equation \( x - x^2 = 0 \). Factor this equation as \( x(1-x) = 0 \). The solutions to this equation are \( x=0 \) and \( x=1 \). Thus, the region of interest is from \( x = 0 \) to \( x = 1 \).
03
Set Up the Integral for Area
The area under the curve \( y = x-x^2 \) from \( x = 0 \) to \( x = 1 \) can be found using the definite integral. We set up the integral as: \[ \int_{0}^{1} (x - x^2) \, dx \].
04
Integrate the Function
Find the integral of \( x - x^2 \):First, integrate \( x \) to get \( \frac{x^2}{2} \), and integrate \( -x^2 \) to get \( -\frac{x^3}{3} \). Thus, the integral becomes:\[ \int (x - x^2) \, dx = \frac{x^2}{2} - \frac{x^3}{3} + C \].
05
Evaluate the Integral from 0 to 1
Evaluate the definite integral from 0 to 1:\[ \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1} = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) \].Calculate:\[ \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \].
06
State the Area of the Region
The area of the region under the curve \( y=x-x^2 \) from \( x=0 \) to \( x=1 \) is \( \frac{1}{6} \) square units.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
Integrals are a fundamental concept in calculus, used to find a variety of quantities such as area, volume, and more. A definite integral is specifically used to compute the net area under a curve over a specified interval on the x-axis.
When we calculate a definite integral, we're looking at the total accumulation of quantities, such as area, between two points. For a function \( f(x) \) over the interval \([a, b]\), the definite integral is represented as \( \int_{a}^{b} f(x) \, dx \). This notation indicates that we need to sum up small slices of area from \( x = a \) to \( x = b \).
In our example, the integral \( \int_{0}^{1} (x - x^2) \, dx \) helps us find the area beneath the curve \(y = x - x^2 \) and the x-axis from 0 to 1.
When we calculate a definite integral, we're looking at the total accumulation of quantities, such as area, between two points. For a function \( f(x) \) over the interval \([a, b]\), the definite integral is represented as \( \int_{a}^{b} f(x) \, dx \). This notation indicates that we need to sum up small slices of area from \( x = a \) to \( x = b \).
- The procedure involves finding antiderivatives of the function over the interval.
- You then evaluate these antiderivatives at the bounds, subtract the lower bound value from the upper.
- This process gives us the net area under the curve between the two points.
In our example, the integral \( \int_{0}^{1} (x - x^2) \, dx \) helps us find the area beneath the curve \(y = x - x^2 \) and the x-axis from 0 to 1.
Area Under a Curve
The concept of the area under a curve refers to the space between the curve of a graphed function and the x-axis within a specified interval. This area is often a representation of some physical quantity, like distance or probability.
Finding the area under a curve is essential in various fields including physics and engineering, where it can represent quantities such as work done by a force or the probability density in statistics. To find this area precisely, especially when the curve is irregular, we use calculus and the concept of integration.
Finding the area under a curve is essential in various fields including physics and engineering, where it can represent quantities such as work done by a force or the probability density in statistics. To find this area precisely, especially when the curve is irregular, we use calculus and the concept of integration.
- The idea is to take an infinite sum of infinitesimally small rectangles beneath the curve.
- This leads to the integral formula, summing these slices gives us the total area.
- For our curve \( y = x - x^2 \), the definite integral from 0 to 1 gives an exact measure of the area under this curve.
Parabola
A parabola is a symmetrical, U-shaped curve that can open upwards or downwards on a graph. Parabolas are crucial in calculus and algebra due to their predictable shape and properties.
The general quadratic equation \( ax^2 + bx + c \) describes a parabola. In our exercise, the equation is \( y = x - x^2 \), which rearranges to \( y = -x^2 + x \). Here:
Understanding these properties allows us to visualize parabolas successfully, ensuring we can set up integrals correctly to find areas under such curves.
The general quadratic equation \( ax^2 + bx + c \) describes a parabola. In our exercise, the equation is \( y = x - x^2 \), which rearranges to \( y = -x^2 + x \). Here:
- The parabola opens downwards because the coefficient of \( x^2 \) is negative.
- It symmetrically spans from its vertex, and intersects the x-axis at the roots.
- Roots of our parabola can be found by solving \( x - x^2 = 0 \), leading to the points \( x = 0 \) and \( x = 1 \).
Understanding these properties allows us to visualize parabolas successfully, ensuring we can set up integrals correctly to find areas under such curves.