(a) Find the number a such that the line x=a bisects the area under the curve y=1 / x^2, 1 ⩽x ⩽4

(a) Find the number a such that the line x=a bisects the...

Key Concepts

Definite Integration

Definite integration is a core concept in calculus that involves computing the accumulation of quantities, such as area, along an interval. It provides a way to exactly determine the net area under a curve by summing infinitely many infinitesimal contributions between two bounds.

Area Under a Curve

The area under a curve is found by taking the definite integral of a function over a specified interval. This concept is fundamental for quantifying the space between the graph of a function and the horizontal axis, and it is widely used in applications ranging from physics to probability.

Area Partitioning

Area partitioning refers to dividing a region into parts that have a specific ratio or equal areas. In calculus, one often sets up an equation where the integral representing one portion of the area is equated to a desired fraction of the total area, and then solves for the unknown dividing parameter.

Solving Equations Involving Integrals

Many calculus problems reduce to solving equations that include definite integrals. These problems require evaluating the integrals, setting them equal to target values derived from geometric or physical constraints, and algebraically manipulating the resulting equation to find the required parameter or value.

Transcript

00:02 Okay, so we have the equation, y equals 1 over x squared, along the interval x must be greater than or equal to 1, and x must be less than or equal to 4.

00:12 And we need to find the value a such that the line x equals a bisects that area underneath the curve, and that curve is drawn out here on the little mini graph.

00:24 Okay, so to solve for a, we are going to use a pair of integrals.

00:29 So we're going to integrate this equation using that interval on the graph.

00:34 So the first equation goes from 1 to a, and i'm going to write out y equals 1 over x squared as y equals x to the negative second power.

00:45 You'll see y in a second, and we'll set that equal to the interval from a to 4, x to the negative second power, just the same.

01:02 All right, so in order to integrate both of these and solve for a, we are going to use the power rule.

01:09 And if you're unfamiliar with the power rule, it's pretty simple.

01:12 I'll outline it here.

01:15 You take the exponent, add one, and then divide by the value of the exponent plus one, like so.

01:29 And then you set it against the interval x equals 1 to x equals a.

01:34 And set that equal to, and i'll write it out a little bit neater here as negative x to the negative first power from x equals a.

01:50 To x equals 4.

01:52 Now what i've written out here is the same thing that appears on the left side, just simplified a little bit, so it's easier.

01:59 Now, for both of these, you are going to plug in these values, and that'll be on this side that looks like x equals a minus the value of x equals 1, and on this side you plug in 4 for x, and then subtract a equals x from that.

02:21 I'll write that all out so it's a little easier to understand.

02:24 So on the slide you'll get negative a to the negative first power minus negative 1 to the negative first power and set that equal to negative 4 to the negative first power minus negative to the negative first power.

02:51 Now at this point, it just becomes basic algebra.

02:54 You're going to solve for a and once you've done that, you will get the answer.

03:02 A is equal to a 8 over.

03:08 That is your value for a, for which the line x equals a, bisects the area under the curve.

03:17 Okay, now there is a second part to this equation.

03:26 We are going to clean all this up and get started on that.

03:34 Okay, so the second part of this equation is basically that we have to find a value b.

03:46 Such that the line b equals y also bisects the area under the curve.

03:55 So here, something like that...