Coefficients can be removed when differentiating, i.e. d/dx 2tanx = 2 d/dx tanx
You should know that d/dx tanx = sec2x (either by differentiating sinx/cosx or by just remembering the derivatives of common trig functions, as it will come in handy).
So the answer is 2sec2x
In the formula for electrostatic forces, the value of k is 8.99 x 109 N m2 C-2
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:
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
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.)
There are several simple to use 'Test Kits' available at most pet shops.
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
∫ xⁿ dx = xn + 1/(n + 1) + c where n is any constant except 0 and -1.
Apply that rule to get:
∫ 5x dx
= 5 ∫ x dx [Factor out the constant]
= 5 ∫ x1 dx [Make note of the exponent for x]
= 5x1 + 1/(1 + 1) + c
= (5/2)x2 + c
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.
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.
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.
It depends:
The derivative with respect to Y is: dX/dY 2Y = 2
The derivative with respect to X is: dY/dX 2Y = 0
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.
Same as any other function - but in the case of a definite integral, you can take advantage of the periodicity. For example, assuming that a certain function has a period of pi, and the value of the definite integral from zero to pi is 2, then the integral from zero to 2 x pi is 4.
Know that ∫eu du = eu du/dx + c
∫-e-x dx = e-x + c
But ∫eu du = eu + c
Perhaps we are integrating -(e-x ) though the question might be (-e)-x
Question is not clear.
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.
Delta represents a change. Therefore, "delta x" means "change in x."
first you divide both sides by 2 then add brackets in front of the number:
divide 8x+12= 4 and 6 divide again = 4 and 2 and 3
add brackets= 4{2x=3}
Hope This Helped
f(x) = 2/x = 2x^-1
so dy/dx = -2x^-2 = -2/x²
f(x) = 2/x
This can be re-written as f(x) = 2x^-1 according the to the laws of exponents. Now, we just take the derivative normally:
f(x) = 2x^-1
f ' (x) = -1 × 2x^(-1-1) = -2x^-2
The above can be re-written in quotient form as f ' (x) = -2/x²
Retardation is the application of a force that produces negative accelleration. Synonyms would be braking, decelleration, damping, etc. Gravitational force operates downward (in a negative direction) so, in most frames of reference, gravity is a retarding force.
Note that (1/x³) + x = x-3 + x
Then, by the power rule, we integrate the expression to get:
x-3 + 1/(-3 + 1) + x1 + 1/(1 + 1) + c
= -x-2/2 + x2/2 + c where c is the arbitrary constant
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