# Calculus

## The branch of mathematics that deals with the study of continuously changing quantities, with the use of limits and the differentiation and integration of functions of one or more variables, is called Calculus. Calculus analyzes aspects of change in processes or systems that can be modeled by functions. The English physicist, Isaac Newton, and the German mathematician, G. W. Leibniz, working independently, developed calculus during the 17th century.

###### Asked in Jobs & Education, Calculus

### What is the indefinite integral of 2x ln2x dx?

In order to work out this problem, we need to learn how to apply
the integration method correctly.
The given expression is ∫ 2xln(2x) dx.
Instead of working out with 2x's, we let u = 2x. Then, du = 2 dx
or du/2 = dx. This method is both valid and easy to avoid working
out with too much expressions. You should get:
∫ uln(u) (du/2)
= ½ ∫ uln(u) du
Use integration by parts, which states that:
∫ f(dg) = fg - ∫ g(df)
We let:
f = ln(u). Then, df = 1/u du
dg = u du. Then, g = ∫ u du = ½u²
Using these substitutions, we now have:
½(½u²ln(u) - ½∫ u du)
= ¼(u²ln(u) - ∫ u du)
Finally, by integration, we obtain:
¼ * (u²ln(u) - ½u²) + c
= 1/8 * (2u²ln(u) - u²) + c
= 1/8 * (2(2x)²ln(u) - (2x)²) + c
= 1/8 * (2x)² * (2ln(u) - 1) + c
= ½ * x² * (2ln(2x - 1)) + c

###### Asked in Calculus, Trigonometry

### How do you solve x equals 2 sin x?

The problem x = 2 sin x cannot be solved by using algebraic
methods.
One solution is to draw the graphs of y = x and y = 2 sin x.
The two lines will intersect. The values of x where the
intersection takes place are the solutions to this problem.
You can tell from the graph that one solution is x=0 and verify
this contention by noting that 2 sin(0) = 0.
You can find the other solution through numerical methods and
there are many that will give you the correct solution. Perhaps the
simplest is to start with a value of X like pi/2 and then take the
average of 2*sin(X) and X. Using that as your new value, again take
the average of 2*sin(X) and X. As you continue to do this, the
value will get closer and closer to the desired value. After 20
steps or so, the precision of your calculator will probably be
reached and you will have a pretty good answer of about
1.89549426703398. (A spreadsheet can be used to make these
calculations pretty easily.)

###### Asked in Math and Arithmetic, Calculus

### What is the antiderivative of -cscxcotx?

First, antiderivative = a solution to the indefinite
integral
therefore to integrate -(csc(x))(cot(x)) first convert it to
-cos(x)/sin2(x)
To integrate ∫-cos(x)/sin2(x) dx, use substitution u = sin(x)
and du/dx = cosx
This will make it ∫-1/u2 du and the antiderivative is 1/u +c,
therefore the answer is 1/sin(x) + c.

###### Asked in Math and Arithmetic, Algebra, Calculus

### If f of x equals 10 plus x squared then f of x and h equals?

If f(x)=x2+10, then f(x+h)=?
f(x+h)=(x+h)2+10 (since f(x)=x2+10, substitute the x in x2 to
(x+h)2)=(x+h)(x+h)+10 (then multiply (x+h) by (x+h) by doing the
FOIL method)=x2+xh+xh+h2+10 (First: x*x, Outside:
x*h, Inside: h*x, Last:
h*h)=x2+2xh+h2+10 (combine like
terms (xh+xh=2xh))
So if f(x)=x2+10, then
f(x+h)=x2+2xh+h2+10

###### Asked in Math and Arithmetic, Algebra, Calculus

### What is the integral of 3x?

In order to compute that integral, we need to use the power
rule:
∫ xⁿ dx = xn + 1/(n + 1) + c where n is any constant except 0
and -1.
Apply that rule to get:
∫ 3x dx
= 3 ∫ x dx [Factor out the constant]
= 3 ∫ x1 dx [Make note of the exponent]
= 3x1 + 1/(1 + 1) + c
= 3x2/2 + c
So that is the integral of 3x.

###### Asked in Statistics, Calculus

### Can the first quartile equal the third quartile?

yes, if all the data is the same number; when the range is
zero.
* * * * *
That is not true. You need 25% of the values to be small, then
50% identical values, followed by 25% large values. Then the lower
(first) quartile will be the same as the upper (third) quartile.
The inter-quartile range (IQR) will be zero but the overall range
can be as large as you like.

###### Asked in Calculus

### What is the derivative of 2Y?

###### Asked in Calculus

### What are some real life applications of calculus?

Pretty much any serious statistical model or experiment on
anything will use basic calculus to interpret data.
Anything that exponentially grows or decays (radioactive matter,
bacteria, population growth, etc.)
Anything that's built to be structurally sound.
Anything that uses the EM spectra (radio, microwaves, visible
light, etc.)
All scientific industries use calculus practically
constantly.
And on and on and on...
In reality, it's rarely pure theoretical calculus that's being
done. Rather, another branch of math based on and built from the
principles and results of calculus is primarily used called
differential equations.
Don't forget integration, the other "half" of calculus. That is
as equally important in your listed applications.
Also, both theoretical and applied calculus use both
differentiation and integration. Differentiation isn't a separate
branch of maths, but one of the two major branches of calculus as a
whole.

###### Asked in Calculus, Geometry

### What is the geometrical interpretation of line integral?

A line integral or a path or curve integral is a function used
when we integrate along a curve. The first step is to parametrize
the curve if it is not already in parametric form. So let's look at
a general curve, c in its parametric form.
The letter t will stand for time.
x=h(t) and y=g(t) so both x and y are functions of time. (We
assume the curve is smooth)
We will integrate over an interval [a,b] so a≤t≤b
Now to understand what it means geometrically, let's look at a
common geometric shape, the circle. x2 +y2 =16 is the equation of a
circle centered at the origin with radius 4. We are going to
integrate xy4 along half of the circle. We first need it in
parametric form. So x=4cos(t) and y=4sin(t). Keep in mind this
circle will be our domain. Just like we find definite integrals
over an interval, we use the curve as our analogous domain.
We are going to integrate over half the circle so we need to
specify t so that our curve will trace out half the circle. We can
let -pi/2≤t≤pi/2
Next we compute ds so we need to find dx/dt and dy/dt.
So dx/dx=-4sint and dy/dt=4cost.
Now we know that ds=Square root (16sin2 t + 16cos2 t)dt=4dt. (
ds is used to show we are moving along the curve)
So the integral of xy4 along a general curve c, becomes the
integral of cos(t)sin4 (t) dt
integrated between -pi/2 to pi/2.
The result of integration along this path is 8192/5 so now we
ask what does this mean geometrically? (Took a bit of time to get
here, but we needed the example)
Imagine putting some points on the half circle curve we used.
Say point 0, 1, 2, 3, 4 etc. Now draw vectors from 0 to 1, 1 to 2,
2 to 3 etc.
We have partitioned our curve into subcurves, the vectors. This
is like dividing an interval into the rectangles you have often
seen for Riemann integrals and Riemann sums, Instead of looking at
the areas of each of those rectangles, we look at the f(pn) where
pn are the points 0,1,2,3 etc mentioned above. Just like we add up
the sum of the areas of the rectangles for a Riemann sum, we add up
the values of the function at each point on our curve, in this case
a half circle.
Now if we take the limit as the vectors get smaller or as the
length of the subintervals along the curve tends toward 0, this is
the line integral
The path is being partitioned into a polygonal path. We are
letting the length or mesh or the partition go to 0. You can do
this in 2 or 3 dimension space and we can integrate over a vector
field.
This last example helps you see an even more geometric example.
Sometimes the geometry is hard to see in the examples like the one
I gave above. In this one, it is pretty easy.
suppose we have a function of two variables that's just a sheet
above the x-y plane. Let's look at f(x,y)=5 and we want to take the
line integral of this function around the closed path which is the
unit circle x2+y2=1. Geometrically, the line integral is the area
of a cylinder of length 1 and radius 5.
You can check the this is 10Pi
( remember the surface area of a cylinder is 2Pirh, which is
2Pi5x1=10Pi)
This one is easier to see geometrically because f(x,y)=5 is a
sheet above the x-y plane which is easy to see. If we evaluate the
that along the unit circle, then we can see how the cylinder would
have to be length or height 1 and radius 5 from the sheet. To tie
this to what we said earlier, think of dividing the path, the unit
circle, into small subpaths or polygonal paths and evaluating
f(x,y)=5 on each of these path. For each one, we get a small piece
of the cylinder. If you have trouble seeing this, consider the
small piece of the circle between t=0 and t=pi/1000
A very small sector. Now when we look at the value of the
function on that very think sector, we have a pie shaped sector
with radius 5. We add up lots of these and get the cylinder.
Line integrals are also commonly used to find the work done by
force long a curve.
One last thing to help understand the geometry. Since we talked
about vectors and dividing the curve into vectors, let's look at
that geometry.
Of course, the dot product of any two vectors is positive if the
point in the same direction and zero if they are perpendicular. It
is negative it they point in roughly opposite directions. The line
integral geometrically add up dot products. If F is a vector field
and DeltaR is a displacement vector. We look at the dot product of
F and DeltaR long the path. If the magnitude of F is constant, the
line integral will be a positive number if F is mostly pointing the
same way as DeltaR and negative if
F is pointing in the opposite direction. If the F is
perpendicular to the path at all times, the line integral has a
value of zero.
We should mention one last thing ( it really is the last
thing)
Imagine a piece of string or wire. Let that be the curve we have
been talking about analogous to the half circle etc. Now suppose we
have a function f(x,y) that tells us the mass of the wire or string
per unit length. The mass of the string if the line integral of
f(x,y) evaluated along the string.
An ordinary everyday integral is of y (a function of x) which is
just a sum of all the values of the thin rectangles y times dx for
all the tiny pieces of dx which make up the x axis. It can be
regarded as a line integral along the line known as the x axis. But
you could do an integral along any other line or curve, provided
you know the value of the function at all points on the line or
curve. And that is called a line integral. Instead of the fragment
dx, you have a fragment ds where s is your position along the line,
just as x describes where you are on the x axis.

###### Asked in Math and Arithmetic, Calculus

### What is the derivative of 2cosx times the inverse of sinx?

d/dx (2cos(x)sin⁻¹(x))
right, 2cos(x)sin⁻¹(x) is a product, so we'll differentiate
using the product rule:
d/dx(uv) = u d/dx(v) + v d/dx(u)
u = 2cos(x)
v = sin⁻¹(x)
d/dx(u) = -2sin(x)
to find d/dx(sin⁻¹(x)) we'll set
y=sin⁻¹(x)
x=sin(y)
dx/dy = cos(y)
dy/dx = 1/(cos(y)) cos²(y) + sin²(y)=1,
dy/dx = 1/sqrt(1-sin²y) = 1/sqrt(1-x²) [x=sin(y)]
so plugging all this into our product rule,
d/dx (2cos(x)sin⁻¹(x))
= 2cos(x)/sqrt(1-x²) - 2sin(x)sin⁻¹(x).

###### Asked in Math and Arithmetic, Physics, Calculus

### Formula for velocity?

V = d / t (or v = dd/dt & v = integral of acceleration with
respect to time for physics involving calculus)
where V is velocity, d is displacement (distance traveled) and t
is time.
vf = vi + a*t
(vf)^2 = (vi)^2 + 2*a*d
Where vf is final velocity, vi is initial velocity, and a is
acceleration
v = p/m
Where p is object's momentum and m is object's mass