Question

1. # Find The Number A Such That The Line X = A Bisects The Area Under The Curve Y = 1/X2 For 1 ≤ X ≤ 4

## Introduction

Finding the area under a curve is a mathematical concept that can be incredibly useful in a number of different fields. It can be used to calculate the area of a triangle, the total length between two points, and more. In this article, we’ll explore how to find the number A such that the line X = A bisects the area under the curve Y = 1/X2 for 1 ≤ X ≤ 4. We’ll also look at how this technique can be used to solve more complex problems. So if you’re looking to brush up on your math skills or learn something new, read on!

## What is the Area Under a Curve?

The area under a curve is the space enclosed between the curve and the x-axis. The most common way to find the area under a curve is to use integration. Integration is a mathematical process that finds the area under a curve by summing up theinfinitesimal rectangles that make up the area.

## How to Find the Number A

To find the number A such that the line X = A bisects the area under the curve Y = /X for ≤ X ≤, we need to set up an equation and solve for A. The first step is to find the area under the curve. This can be done by using calculus or by finding a numerical approximation. For this particular problem, we will use calculus.

The next step is to set up the equation that we will need to solve. We know that we want the line X = A to bisect the area under the curve, so our equation will be:

A= ∫YdX

We can solve this equation for A by using integration by parts. First, we will take the derivative of both sides of the equation:

dA/dX= d/dX(∫YdX)
= d/dX(Y*X- ∫X*dY)
= Y + d/dX(- ∫X*dY)
= Y – X*(dY/dX)
= Y – X*(-1/X^2)
= Y + 1/X^2

## The Bisection Method

There are a few different ways to find the number A such that the line X = A bisects the area under the curve Y = /X for ≤ X ≤ . One method is the bisection method.

To use the bisection method, we need to first find two numbers that bracket the root, meaning that one number is less than the root and one number is greater than the root. We can then take the midpoint of these two numbers and see if it is greater than or less than the root. If it is less than the root, we know that our new lower bound is this midpoint. If it is greater than the root, we know that our new upper bound is this midpoint. We can then continue taking midpoints until we get close enough to the root.

For example, say we want to find A such that X = A bisects the area under Y = /X for 2 ≤ X ≤ 3 . We can start by finding two numbers that bracket the root: 2 and 3 . The midpoint of these two numbers is 2.5 . Is 2.5 < 3 ? Yes, so our new lower bound becomes 2.5 . We can then take the midpoint of 2.5 and 3 , which is 2.75 . Is 2.75 < 3 ? No, so our new upper bound becomes 2.75 . We can continue taking midpoints in this way until we get close enough to 3.

## Conclusion

We have presented a mathematical approach to finding the number A such that the line X = A bisects the area under the curve Y = 1/X2 for 1 ≤ X ≤ 4. By setting up an integral equation and solving it, we were able to find A which was equal to 2.5. We can now conclude that the number 2.5 is the solution of our problem and it satisfies all conditions stated in this article.

2. Have you ever been presented with a problem involving finding a number, such that a line bisects the area under a given curve? If so, today we’ll be looking at how to find the number ‘A’ such that the line x = A bisects the area under the curve y = 1/x2 for 1 ≤ x ≤ 4.

Let’s begin by breaking down the problem into smaller components. Firstly, we need to identify the curve, y = 1/x2. This curve is known as a hyperbola and is symmetric about the origin (0,0). The area under this curve from x = 1 to x = 4 can be found using integration.

Integrating from x = 1 to x = 4, we get the area under the curve, A = 1/3 – 1/16.

Now, we need to find the line x = A such that it bisects this area. This can be done by finding the x-intercept of the line, which is the point (A, 0). By setting up an equation, we get A = 1/3 – 1/16 = 7/48.

Therefore, the number A such that the line x = A bisects the area under the curve y = 1/x2 for 1 ≤ x ≤ 4 is 7/48.

Now that we’ve found the answer to this question, you can use the same approach in a variety of other problems which involve finding a line which bisects a given area.