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Mobile App Chapter 4: Problem 3
Berechnen Sie die Fl?che zwischen der Parabel \(y=x^{2}\) und der Geraden \(y=-x+6\)
Short Answer
Expert verified
The area is approximately 23.83.
Step by step solution
01
Identify Intersection Points
Set the two equations equal to each other to find the points where the parabola and the line intersect: \[ x^2 = -x + 6 \] Solve for x.
02
Solve the Quadratic Equation
Rearrange the equation to standard form and solve for x: \[ x^2 + x - 6 = 0 \] Factorize or use the quadratic formula: \[ x = \frac{-1 \pm \sqrt{1 + 24}}{2} \] This gives us the solutions: \[ x = 2 \quad \text{and} \quad x = -3 \]
03
Set Up the Integral
The area between the curves from the intersection points can be found using: \[ \int_{-3}^{2} [(-x + 6) - x^2] \, dx \] This represents the integral of the difference between the line and the parabola.
04
Integrate
Evaluate the integral: \[ \int_{-3}^{2} (-x + 6 - x^2) \, dx \] Find the antiderivative: \[ \left[ -\frac{x^2}{2} + 6x - \frac{x^3}{3} \right]_{-3}^{2} \]
05
Calculate the Definite Integral
Substitute the limits into the antiderivative: \[ \left[-\frac{2^2}{2} + 6(2) - \frac{2^3}{3} \right] - \left[-\frac{(-3)^2}{2} + 6(-3) - \frac{(-3)^3}{3} \right] \] Compute each term: \[ \left[-2 + 12 - \frac{8}{3} \right] - \left[-\frac{9}{2} - 18 + 9 \right] \]
06
Simplify the Expression
Simplify the values inside the brackets: \[ \left[10 - \frac{8}{3} \right] - \left[ -\frac{9}{2} - 9 \right] \] Combined and simplified as: \[ \frac{30}{3} - \frac{8}{3} + \frac{15}{2} + 9 \] = \[ \frac{22}{3} + \frac{15}{2} + 9 \]
07
Find the Final Answer
Combine the fractions and whole number to get the final area. Converting everything into the same common denominator and adding: \[ \frac{44}{6} + \frac{45}{6} + \frac{54}{6} = \frac{143}{6} = 23.83 \] The area is approximately \(23.83\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
parabola
A parabola is a U-shaped curve that can open up, down, left, or right. It is described by a quadratic function of the form \(y = ax^2 + bx + c\), where 'a', 'b', and 'c' are constants. The simplest form is when \(a = 1, b = 0,\) and \(c = 0\), making \(y = x^2\).
The vertex of a parabola is its highest or lowest point, depending on its orientation. For \(y = x^2\), the vertex is at the origin (0,0) and the parabola opens upwards.
In our exercise, the parabola is described by the function \(y = x^2\). It intersects the line at specific points, helping us find the area between them using calculus.
The vertex of a parabola is its highest or lowest point, depending on its orientation. For \(y = x^2\), the vertex is at the origin (0,0) and the parabola opens upwards.
In our exercise, the parabola is described by the function \(y = x^2\). It intersects the line at specific points, helping us find the area between them using calculus.
line intersection
Finding the intersection points of a line and a curve like a parabola involves setting their equations equal to one another.
For our exercise, we set \(y = x^2\) and \(y = -x + 6\). By solving \(x^2 = -x + 6\), we find where these two graphs cross.
After rearranging to \(x^2 + x - 6 = 0\), we solve for x. This gives us the points of intersection, crucial for setting up our integral.
Here, the solutions are \(x = 2\) and \(x = -3\). These points tell us the limits for our definite integral, defining the region's boundaries.
For our exercise, we set \(y = x^2\) and \(y = -x + 6\). By solving \(x^2 = -x + 6\), we find where these two graphs cross.
After rearranging to \(x^2 + x - 6 = 0\), we solve for x. This gives us the points of intersection, crucial for setting up our integral.
Here, the solutions are \(x = 2\) and \(x = -3\). These points tell us the limits for our definite integral, defining the region's boundaries.
definite integral
A definite integral calculates the area under a curve between two bounds. In our case, it's between the intersection points we found.
The integral \(\int_{-3}^{2} [(-x + 6) - x^2] \, dx\) subtracts the parabola function from the line function over the interval from -3 to 2, representing the area between these curves.
We use the antiderivative to solve this integral. Integrating the expression inside gives us the function to evaluate at our bounds: \(\left[ -\frac{x^2}{2} + 6x - \frac{x^3}{3} \right]_{-3}^{2}\), which yields the area value after substituting and computing the definite integral.
The integral \(\int_{-3}^{2} [(-x + 6) - x^2] \, dx\) subtracts the parabola function from the line function over the interval from -3 to 2, representing the area between these curves.
We use the antiderivative to solve this integral. Integrating the expression inside gives us the function to evaluate at our bounds: \(\left[ -\frac{x^2}{2} + 6x - \frac{x^3}{3} \right]_{-3}^{2}\), which yields the area value after substituting and computing the definite integral.
quadratic equation
A quadratic equation takes the form \(ax^2 + bx + c = 0\), where 'a', 'b', and 'c' are constants. Solving it gives us the values of x that satisfy the equation.
Common methods include factoring, completing the square, or using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
In our exercise, we solved \(x^2 + x - 6 = 0\) to find where the parabola \(y = x^2\) intersects the line \(y = -x + 6\). The solutions were \(x = 2\) and \(x = -3\), crucial for our area calculation as they set the limits for our integral.
Common methods include factoring, completing the square, or using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
In our exercise, we solved \(x^2 + x - 6 = 0\) to find where the parabola \(y = x^2\) intersects the line \(y = -x + 6\). The solutions were \(x = 2\) and \(x = -3\), crucial for our area calculation as they set the limits for our integral.
