Evaluate the function
ƒ(n) = n
52at ƒ(0)
Given ƒ(n) = n
52Determine the derivative ∫ƒ(n)
Given ƒ(n) = n
52Determine the 2nd derivative ∫ƒ(n)
Given ƒ(n) = n
52dn
Determine the integral ∫ƒ(n)
Go through and integrate each term
Given the equation ƒ(n) = n
52use Simpsons Rule with n =
over the interval [0,1]
Substitute each n with 0
ƒ(
0) = (
0)
52ƒ(
0) = (0)
ƒ(
0) = 0
Start ƒ'(n)
Use the power rule
ƒ'(n) of an
n = (a * n)n
(n - 1)For this term, a = 1, n = 52
and n is the variable we derive
ƒ'(n) = n
52ƒ'(n)( = 1 * 52)n
(52 - 1)ƒ'(n) = 52n
51
Collecting all of our derivative terms
ƒ'(n) =
52n51Start ƒ''(n)
Use the power rule
ƒ''(n) of an
n = (a * n)n
(n - 1)For this term, a = 52, n = 51
and n is the variable we derive
ƒ''(n) = 52n
51ƒ''(n)( = 52 * 51)n
(51 - 1)ƒ''(n) = 2652n
50
Collecting all of our derivative terms
ƒ''(n) =
2652n50Start ƒ(3)(n)
Use the power rule
ƒ
(3)(n) of an
n = (a * n)n
(n - 1)For this term, a = 2652, n = 50
and n is the variable we derive
ƒ
(3)(n) = 2652n
50ƒ
(3)(n)( = 2652 * 50)n
(50 - 1)ƒ
(3)(n) = 132600n
49
Collecting all of our derivative terms
ƒ
(3)(n) =
132600n49Start ƒ(4)(n)
Use the power rule
ƒ
(4)(n) of an
n = (a * n)n
(n - 1)For this term, a = 132600, n = 49
and n is the variable we derive
ƒ
(4)(n) = 132600n
49ƒ
(4)(n)( = 132600 * 49)n
(49 - 1)ƒ
(4)(n) = 6497400n
48
Collecting all of our derivative terms
ƒ
(4)(n) =
6497400n48Evaluate ƒ(4)(0)
ƒ
(4)(0) = 6497400(
0)
48ƒ
(4)(0) = 6497400(0)
ƒ
(4)(0) = 0
Given ƒ(n) = n
52dn
Determine the integral ∫ƒ(n)
Go through and integrate each term
Integrate term 1
ƒ(n) = 6497400n
48Use the power rule
∫ƒ(n) of the expression an
n = 6497400, n = 48
and n is the variable we integrate
∫ƒ(n) = | 6497400n(48 + 1) |
| 48 + 1 |
Simplify our fraction.
Divide top and bottom by 49
∫ƒ(n) = 132600n
49
Collecting all of our integrated terms we get:
∫ƒ(n) =
6497400n49132600n49Evaluate ∫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) =
132600Evaluate ∫ƒ(0)
∫ƒ(0) = 6497400(
0)
49132600(
0)
49∫ƒ(0) = 6497400(0)132600(0)
∫ƒ(0) = 0
∫ƒ(0) =
0Determine 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)
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