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]

ƒ(0) = (0)52
ƒ(0) = (0)
ƒ(0) = 0

## 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

ƒ'(n) = 52n51

## 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

ƒ''(n) = 2652n50

## 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

## 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

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

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

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