Featured 3 months ago

=

=

and then

=

Featured 3 months ago

Please see below.

Every point on the curve has coordinates

We can minimize the distance by minimizing the radicand:

Differentiate:

Use some technology or approximation method to get

There cannot be a maximum. There is a minimum at this

Using the distance above, we find a minimum distance of approximately

Featured 3 months ago

See below.

The critical points are the points where the first derivative equals zero, or the point doesn’t exist.

This is a polynomial so it is continuous for all

First derivative of

Equating this to zero:

Factor:

Putting these values in

Critical Points:

Increasing in

Decreasing in

GRAPH:

Featured 3 months ago

A few thoughts...

**Definite vs indefinite integral**

A definite integral includes a specification of the set of values over which the integral should be calculated. As a result, it has a definite value, e.g. the area under a curve in a given interval.

By way of contrast, an indefinite integral does not specify the set of values over which the integral should be calculated. It basically identifies what the antiderivative function looks like, including some constant of integration to be determined. For example:

#int x^2 dx = 1/3 x^3 + C#

**Non-elementary integrals of elementary functions**

Unlike derivatives, the integral of an elementary function is not necessarily elementary. The term "elementary function" denotes functions constructed using basic arithmetic operations,

There are some very useful non-elementary functions expressible as integrals of elementary functions. For example, the Gamma function:

#Gamma(x) = int_0^oo t^(x-1) e^(-t) dt#

The Gamma function extends the definition of factorial to values apart from non-negative integers.

**Poles and Cauchy principal value**

If a function has a singularity such as a simple pole, then its definite integral over a range including that pole is not automatically well defined. A workaround for such cases is provided by the Cauchy principal value.

For example:

#int_(-1)^1 dt/t = lim_(epsilon -> 0+) (int_(-1)^-epsilon dt/t + int_epsilon^1 dt/t) = 0#

**Non measurable sets**

If the set over which you are trying to integrate is non-measurable, then the integral is usually not defined. An exception would be if the value of the function on that set was zero.

To 'construct' a non-measurable set you would typically use the axiom of choice.

For example, you could define an equivalence relation on

#a ~ b <=> (a-b) " is rational"#

This equivalence relation partitions

Use the axiom of choice to choose exactly one element of each equivalence class to create a subset

For any rational number

We can define a non-integrable function by:

#f(t) = { (1 " if " t in S_x " where " x = p/q " in lowest terms and " q " is even"), (0 " otherwise") :}#

This function is not integrable over any interval.

Featured 3 months ago

We have

Now, let's try to remember the definition of an integral. This might seem strange, but it will come of great use.

The definite integral

This concept is called a

There are infinitely many rectangles, therefore, if there are

Now, let's take the case where the width is constant between all boxes. If we define

The Riemann Sum aproximates an integral by being the sum of all the rectangles. In order to simplify this further, let's take the particular case where

This is only true if

One way to make

This looks familiar to our original sum,

All we are left to do is to assume they're equal and then solve the integral.

We have to find the integral of this function:

Since

This is because it forms a rectangle with the x-axis, the width and lenght of which is

As a conclusion,

Featured 2 months ago

We see that

Both the numerator and denominator are a sum or difference of cubes, for which we have a formula:

Thus, we have

We can group

Let the product in

Notice how some things cancel each other:

It might seem confusing, but the numerator of the

Let's write out

Again, lots of things cancel out.

Therefore, we have:

Now, let's take the

As

In other words, it approaches

Featured 2 months ago

I have the graphs of the functions and lines here:

We want the area of the green region.

We can think of the situation like this:

Now, let's find the intercept and the intersection.

The intercept:

The intersection:

We can now form a rectangle like the following:

The area of the green region is the area of the rectangle minus the area under the curve of

The fundamental theorem of calculus:

Remember that

In case you are wondering how I solved

Featured 2 months ago

Diverges

Many students fall into a trap when solving problems similar to this coming up with an answer of

However, let us begin by noting that this integral does not follow our formal definition by which the integral is continuous on the given interval. Note that

If this original integral did not contain the value

However in this case we are allow to set limits for the integral. From the limits we set it allows the user to determine if the integral approaches a value or diverges. In this case divergency meaning going off in two opposite directions towards

Let us start by computing the integral if we were on a different interval that did not include

Now in this case let us define a new integral

Now since we experience a *bad spot* at

In this case let us define

Let us then find

let us now note that

Remember that

Now from summing and replacing our limits that:

since

We now have an indefinite form. In this case

Featured 2 months ago

The solid from revolving around the

.

This is a parabola that opens down and crosses the

If this area is revolved about the

If we cut a vertical slice through the solid with a thickness

The area of this circular disc is:

Then the volume of this disc would br:

Now, imagine that the thickness

If we take the integral of the area of this disc and evaluate it from

If the area between the parabola and the

If we cut a vertical slice through this solid with a thickness of

We can calculate the area of the washer by subtracting the area of the inner circle from the area of the outer circle:

The volume of the washer is:

Now, imagine that the thickness of the washer

To do this, we will take the integral of the area function:

Therefore, the solid from revolving around the

Featured 1 month ago

There is no any trig.function in the answer .So it is difficult to proceed with trig substitution.

Here,

Let,

Now,

Take,

Subst .back

Hence, from