##### Questions

##### Question type

Use these controls to find questions to answer

Featured 4 months ago

Please refer to the explanation below.

We have to take the derivative of the function

So, we have to use the quotient rule, which states that,

Here,

Plugging in

So, the instantaneous rate of change will be

Featured 3 months ago

Two lines are parallel if they share the same slope.

Any line parallel to

Now, we have:

We try to find

=>

We use the power rule:

=>

Now, we set

=>

=>

=>

We plug this value in to get our

=>

=>

Therefore, the point is at

Featured 3 months ago

Use the substitution

Let

#I=int1/sqrt(4x+8sqrtx-24)dx#

Complete the square in the square root:

#I=1/2int1/sqrt((sqrtx+1)^2-7)dx#

Apply the substitution

#I=int(sqrt7u-1)/sqrt(u^2-1)du#

Rearrange:

#I=sqrt7intu/sqrt(u^2-1)du-int1/sqrt(u^2-1)du#

Apply the substitution

#I=sqrt7sqrt(u^2-1)-intsecthetad theta#

Hence

#I=sqrt(7u^2-7)-ln|sectheta+tantheta|+C#

Reverse the last substitution:

#I=sqrt(7u^2-7)-ln|u+sqrt(u^2-1)|+C#

Reverse the first substitution:

#I=sqrt((sqrtx+1)^2-7)-ln|(sqrtx+1)+sqrt((sqrtx+1)^2-7)|+C#

Simplify:

#I=sqrt(x^2+2sqrtx-6)-ln|(sqrtx+1)+sqrt(x^2+2sqrtx-6)|+C#

Featured 3 months ago

Integral is

Let

=

=

Note that when

Now observe the graph of

graph{(x-3)^3 [0.542, 5.542, -1.2, 1.3]}

Note that area of curve in the interval

Featured 2 months ago

See below.

The critical points are the points where the first derivative equals zero, or the point doesnâ€™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 2 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,

Ask a question
Filters

Ã—

Use these controls to find questions to answer

Unanswered

Need double-checking

Practice problems

Conceptual questions