Enter your expression below:


ƒ()
ƒ'()

n =

Evaluate the function
ƒ(n) = n52
at ƒ(0)
Given ƒ(n) = n52
Determine the derivative ∫ƒ(n)
Given ƒ(n) = n52
Determine the 2nd derivative ∫ƒ(n)
Given ƒ(n) = n52dn
Determine the integral ∫ƒ(n)
Go through and integrate each term
Given the equation ƒ(n) = n52
use Simpsons Rule with n =
over the interval [0,1]

Substitute each n with 0

ƒ(0) = (0)52
ƒ(0) = (0)
ƒ(0) = 0

Start ƒ'(n)


Use the power rule

ƒ'(n) of ann = (a * n)n(n - 1)
For this term, a = 1, n = 52
and n is the variable we derive
ƒ'(n) = n52
ƒ'(n)( = 1 * 52)n(52 - 1)
ƒ'(n) = 52n51

Collecting all of our derivative terms

ƒ'(n) = 52n51

Start ƒ''(n)


Use the power rule

ƒ''(n) of ann = (a * n)n(n - 1)
For this term, a = 52, n = 51
and n is the variable we derive
ƒ''(n) = 52n51
ƒ''(n)( = 52 * 51)n(51 - 1)
ƒ''(n) = 2652n50

Collecting all of our derivative terms

ƒ''(n) = 2652n50

Start ƒ(3)(n)


Use the power rule

ƒ(3)(n) of ann = (a * n)n(n - 1)
For this term, a = 2652, n = 50
and n is the variable we derive
ƒ(3)(n) = 2652n50
ƒ(3)(n)( = 2652 * 50)n(50 - 1)
ƒ(3)(n) = 132600n49

Collecting all of our derivative terms

ƒ(3)(n) = 132600n49

Start ƒ(4)(n)


Use the power rule

ƒ(4)(n) of ann = (a * n)n(n - 1)
For this term, a = 132600, n = 49
and n is the variable we derive
ƒ(4)(n) = 132600n49
ƒ(4)(n)( = 132600 * 49)n(49 - 1)
ƒ(4)(n) = 6497400n48

Collecting all of our derivative terms

ƒ(4)(n) = 6497400n48

Evaluate ƒ(4)(0)

ƒ(4)(0) = 6497400(0)48
ƒ(4)(0) = 6497400(0)
ƒ(4)(0) = 0
itarget = 0,1
Given ƒ(n) = n52dn
Determine the integral ∫ƒ(n)
Go through and integrate each term

Integrate term 1

ƒ(n) = 6497400n48

Use the power rule

∫ƒ(n) of the expression ann
an(n + 1)
n + 1

= 6497400, n = 48
and n is the variable we integrate
∫ƒ(n)  =  6497400n(48 + 1)
  48 + 1

∫ƒ(n)  =  6497400n49
  49

Simplify our fraction.
Divide top and bottom by 49
∫ƒ(n) = 132600n49

Collecting all of our integrated terms we get:

∫ƒ(n) = 6497400n49132600n49

Evaluate ∫ƒ(n) on the interval [0,1]

The value of the integral over an interval is ∫ƒ(1) - ∫ƒ(0)

Evaluate ∫ƒ(1)

∫ƒ(1) = 6497400(1)49132600(1)49
∫ƒ(1) = 6497400(1)132600(1)
∫ƒ(1) = 132600
∫ƒ(1) = 132600

Evaluate ∫ƒ(0)

∫ƒ(0) = 6497400(0)49132600(0)49
∫ƒ(0) = 6497400(0)132600(0)
∫ƒ(0) = 0
∫ƒ(0) = 0

Determine our answer

∫ƒ(n) on the interval [0,1] = ∫ƒ(1) - ∫ƒ(0)
∫ƒ(n) on the interval [0,1] = 132600 - 0

Final Answer

ƒ(0) = 0
ƒ(4)(0) = 0
∫ƒ(n) on the interval [0,1] = 132600





What is the Answer?
ƒ(0) = 0
ƒ(4)(0) = 0
∫ƒ(n) on the interval [0,1] = 132600
How does the Functions-Derivatives-Integrals Calculator work?
Free Functions-Derivatives-Integrals Calculator - Given a polynomial expression, this calculator evaluates the following items:
1) Functions ƒ(x).  Your expression will also be evaluated at a point, i.e., ƒ(1)
2) 1st Derivative ƒ‘(x)  The derivative of your expression will also be evaluated at a point, i.e., ƒ‘(1)
3) 2nd Derivative ƒ‘‘(x)  The second derivative of your expression will be also evaluated at a point, i.e., ƒ‘‘(1)
4)  Integrals ∫ƒ(x)  The integral of your expression will also be evaluated on an interval, i.e., [0,1]
5) Using Simpsons Rule, the calculator will estimate the value of ≈ ∫ƒ(x) over an interval, i.e., [0,1]
This calculator has 7 inputs.

What 1 formula is used for the Functions-Derivatives-Integrals Calculator?

Power Rule: f(x) = xn, f‘(x) = nx(n - 1)

For more math formulas, check out our Formula Dossier

What 8 concepts are covered in the Functions-Derivatives-Integrals Calculator?

derivative
rate at which the value y of the function changes with respect to the change of the variable x
exponent
The power to raise a number
function
relation between a set of inputs and permissible outputs
ƒ(x)
functions-derivatives-integrals
integral
a mathematical object that can be interpreted as an area or a generalization of area
point
an exact location in the space, and has no length, width, or thickness
polynomial
an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s).
power
how many times to use the number in a multiplication
Example calculations for the Functions-Derivatives-Integrals Calculator

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