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Find the area bounded by \(y=2-x^{2}\) and \(y=x\).

Short Answer

Expert verified
So, the area bounded by \(y=2-x^{2}\) and \(y=x\) is \(7/3\) square units.

Step by step solution

01

- Finding Points of Intersection

The points of intersection of the two curves can be found by setting \(y=2-x^{2}\) and \(y=x\) equal to each other:\(2-x^{2} = x \)This gives us a quadratic equation that can be solved for \(x\). Rearrange the terms:\(x^{2} + x - 2 = 0 \)This factors to:\( (x-1)(x+2) = 0 \)Setting each factor equal to zero gives the solutions \(x=1\) and \(x=-2\). Plugging these back into \(y=x\) gives the corresponding y-values, \(y=1\) and \(y=-2\). So the points of intersection are \((1, 1)\) and \((-2, -2)\).
02

- Setting Up the Integral

The integral for the area bounded by the two curves from \(x=-2\) to \(x=1\) is given by:\[ \int_{-2}^{1} (2-x^{2}) - x dx \]This expression is obtained by subtracting the equation of the lower curve (\(y=x\)) from the equation of the upper curve (\(y=2-x^{2}\)).
03

- Evaluating the Integral

Now, compute the integral:\[ = -(x^{3}/3) + x^{2} + x -2 | _{-2}^{1} \]Substituting the lower and upper limits of the integral, we compute:\[ = -(1/3) + 1 + 1 - 2 - [(-8/3) + 4 +2 -2 ] = 11/3 -4/3 = 7/3 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental tool in calculus, especially useful when determining areas bounded by curves. When you have two functions that intersect and you want to find the area between them, integration helps you do just that. To find the area, you must integrate the difference between the two functions over a particular interval where they intersect.

Here’s a simpler breakdown:
  • First, identify the two functions or curves involved.
  • Next, determine the interval where these curves intersect.
  • Set up the integral by subtracting the function that lies below from the one that lies above, across the intersection interval.
The definite integral will give us the net area between the curves over the specified interval. In our exercise, we subtracted the curve represented by the linear function from the quadratic function and integrated from -2 to 1.

Remember: If you get a negative area, this likely means that the upper and lower curves were switched in your integral set-up. Double-check your functions to ensure they are in the correct order.
Quadratic Equations
Quadratic equations take the form of x^2 + bx + c = 0, and are crucial in finding where curves intersect, especially when one of them is a parabola. The standard approach to solve for the variable, x, in these equations is to factorize (if possible), use the quadratic formula, or complete the square.

In our example, we set the two equations equal to each other to represent their intersection, leading to a quadratic equation: x^2 + x - 2 = 0. This specific equation is relatively simple and can be factored easily since it decomposes into two binomials: (x - 1)(x + 2) = 0.

With factoring, we set each factor equal to zero. This gives solutions x = 1 and x = -2, which are the x-coordinates of the intersection points.

Quadratic equations are powerful in analyzing and predicting the behavior of parabolas and other curves, especially in understanding their points of intersection with other lines or curves.
Points of Intersection
Finding the points of intersection between two curves is a commonly encountered problem in calculus. This step is indispensable when determining areas between curves. These points are essential as they define the interval over which you will integrate.

The process involves:
  • Equating the two expressions or equations of the curves.
  • Solving this resultant equation, often quadratic, to find the x-values for the points of intersection.
In our example, setting y = 2 - x^2 equal to y = x allowed us to solve for x-values by forming a quadratic equation. The solutions, x = 1 and x = -2, were then used to find corresponding y-values, yielding the full intersection points (1, 1) and (-2, -2).

These intersection points not only define the limits of integration but also provide necessary insight into the geometry of the interacting curves.

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Most popular questions from this chapter

A beam of light is shining onto a screen creating a disk of radius 50 centimeters. The intensity of light is brightest at the center and diminishes away from the center. If the intensity of light at a distance \(r\) from the center of the beam is given by \(f(r)=\frac{150}{20+r^{2}}\) watt/square \(\mathrm{cm}\), find the total wattage of the beam's image on the screen.

Traditionally, when a college football team seems certain to receive a bid to play in the post-season Orange Bowl, fans begin to throw oranges onto the field. Suppose that at one point during a game, the number of oranges per square yard between the goal line and the 30 -yard line is given by \(\rho(x)=\frac{30-x}{3}\) oranges per square yard, where \(x\) is the number of yards from the goal line. If the field is 160 feet ( \(160 / 3\) yards) wide, how many oranges lie between the goal line and the 30 -yard line?

A city is in the shape of a rectangle 4 miles wide by 6 miles long. A river runs through the middle of the city, parallel to the 6 mile-long sides. People prefer to live nearer the water, so the density of people is given by \(\rho(x)=10,000-800 x\) people per square mile, where \(x\) is the distance from the river. (You may ignore the width of the river in this problem.) (a) Show in a sketch how you will need to slice up the region. (b) What is the area of the \(i\) th slice? (c) What is the approximate population in the \(i\) th slice? (d) Write a Riemann sum to estimate the total population of the city. (e) Calculate the exact population by taking the limit of the Riemann sum and evaluating the resulting definite integral.

(a) Suppose that the density of organisms in a certain petridish varies with the distance from the center of the dish. The density at a distance \(x\) centimeters from the center is given by \(f(x)\) organisms per square centimeter. The petri dish is 18 centimeters in diameter. i. Write an integral that gives the number of organisms in the dish. ii. Find the number of organisms in the dish if \(f(x)=100 e^{-x^{2}}\) organisms per square centimeter. (b) Suppose that the density of organisms in a certain petri dish varies with the distance from a strip of nutrients running along the diameter of the dish. The density at a distance \(x\) centimeters from the line of nutrients is given by \(f(x)\) organisms per square centimeter. The petri dish is 18 centimeters in diameter. i. How will you slice up the petri dish? ii. Approximate the number of organisms in the \(i\) th slice. iii. Write a Riemann sum approximating the total number of organisms in the petri dish. iv. Write an integral that gives the number of organisms in the dish.

Consider a box of cereal with raisins. The box is 5 centimeters deep, 25 centimeters tall, and 16 centimeters wide. The raisins tend to fall toward the bottom; assume their density is given by \(\rho(h)=\frac{4}{h+10}\) raisins per cubic centimeter, where \(h\) is the height above the bottom of the box. How many raisins are in the box?

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