Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 5: Problem 49
Sketch the curve and find the total area between the curve and the given interval on the \(x\) -axis. $$ y=x^{2}-x ;[0,2] $$
Short Answer
Expert verified
The total area is \( \frac{2}{3} \) square units.
Step by step solution
01
Understand the Function
The function given is a quadratic function: \( y = x^2 - x \). This is a parabola that opens upwards, as the coefficient of \( x^2 \) is positive.
02
Determine the Behavior of the Function
To understand the behavior of the function over the interval \([0,2]\), evaluate the function at the endpoints. At \( x = 0 \), \( y = 0^2 - 0 = 0 \). At \( x = 2 \), \( y = 2^2 - 2 = 2 \). This tells us that the curve starts at the origin and ends at \( (2, 2) \).
03
Find the Vertex
The vertex form of a parabola \( y = ax^2 + bx + c \) can be used to find the vertex. The formula for the x-coordinate of the vertex is \( x = -\frac{b}{2a} \). Here \( a = 1 \) and \( b = -1 \), so \( x = \frac{1}{2} \). Calculate \( y \) using this value: \( y = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = -\frac{1}{4} \). The vertex is \( \left( \frac{1}{2}, -\frac{1}{4} \right) \).
04
Plot Key Points and Sketch the Curve
Plot the points \( (0,0) \), \( \left( \frac{1}{2}, -\frac{1}{4} \right) \), and \( (2,2) \). The parabola starts at the origin, dips down to its vertex, and then rises back up to \( (2,2) \). Draw a smooth curve through these points to sketch the curve.
05
Set Up the Integral for Area Calculation
The area between the curve \( y = x^2 - x \) and the x-axis from \( x = 0 \) to \( x = 2 \) is given by the definite integral: \[ A = \int_0^2 (x^2 - x) \, dx. \]
06
Evaluate the Integral
To find the integral, solve: \[ \int_0^2 (x^2 - x) \, dx = \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_0^2. \] Evaluate at the bounds: \[ \left( \frac{8}{3} - 2 \right) - \left( 0 \right) = \frac{8}{3} - \frac{6}{3} = \frac{2}{3}. \]
07
Verify the Result
Confirm the calculations by revisiting each step. Re-evaluate if necessary to ensure accuracy in the integral setup and solution.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Functions
A quadratic function has the form \( y = ax^2 + bx + c \). These functions create a curve known as a parabola when graphed. The shape depends heavily on the value of \( a \).
- If \( a > 0 \), the parabola opens upwards, creating a 'U' shape.- If \( a < 0 \), it opens downwards, forming an upside-down 'U'.
In our exercise, the function \( y = x^2 - x \) shows a positive \( a \) (which is 1), indicating an upward opening parabola. To fully recognize the curve's behavior, one often needs to find its vertex and intercepts.
The vertex formula \( x = -\frac{b}{2a} \) helps locate the 'turning point' or vertex of the parabola. For our function, \( a = 1 \) and \( b = -1 \), so the vertex is at \( \left( \frac{1}{2}, -\frac{1}{4} \right) \). This is where the parabola reaches its lowest point before rising again.
- If \( a > 0 \), the parabola opens upwards, creating a 'U' shape.- If \( a < 0 \), it opens downwards, forming an upside-down 'U'.
In our exercise, the function \( y = x^2 - x \) shows a positive \( a \) (which is 1), indicating an upward opening parabola. To fully recognize the curve's behavior, one often needs to find its vertex and intercepts.
The vertex formula \( x = -\frac{b}{2a} \) helps locate the 'turning point' or vertex of the parabola. For our function, \( a = 1 \) and \( b = -1 \), so the vertex is at \( \left( \frac{1}{2}, -\frac{1}{4} \right) \). This is where the parabola reaches its lowest point before rising again.
Definite Integral
The definite integral is a fundamental concept in calculus used to find the area under a curve within a certain interval. It represents the accumulation of a quantity such as area.
For a function \( f(x) \), the definite integral from \( a \) to \( b \) is given by:
\[ \int_a^b f(x) \, dx \]
This notation means you're evaluating the integral from \( x = a \) to \( x = b \). The limits of integration \( a \) and \( b \) define the portion of the curve that you're interested in.
In our exercise, the interval \([0,2]\) is used. We set up the integral:
\[ A = \int_0^2 (x^2 - x) \, dx \]
This integral helps us find how much area is trapped between the curve \( y = x^2 - x \) and the x-axis within these bounds.
For a function \( f(x) \), the definite integral from \( a \) to \( b \) is given by:
\[ \int_a^b f(x) \, dx \]
This notation means you're evaluating the integral from \( x = a \) to \( x = b \). The limits of integration \( a \) and \( b \) define the portion of the curve that you're interested in.
In our exercise, the interval \([0,2]\) is used. We set up the integral:
\[ A = \int_0^2 (x^2 - x) \, dx \]
This integral helps us find how much area is trapped between the curve \( y = x^2 - x \) and the x-axis within these bounds.
Area under a Curve
Finding the area under a curve is an important application of integrals in calculus. The "area under the curve" between the function and the x-axis provides insights into various physical quantities.
To calculate this area, we set up a definite integral using the function. This area can sometimes represent practical things like distance or probability.
To calculate this area, we set up a definite integral using the function. This area can sometimes represent practical things like distance or probability.
- For the function \( y = x^2 - x \) over \([0,2]\), the area is obtained by calculating the integral.
- As seen, the integrated result is \( \frac{2}{3} \). This represents the total area confined by the parabola from \( x = 0 \) to \( x = 2 \) and above the x-axis.
