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Basics of Integration

Basics of Integration

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Fundamental Theorem of Calculus

For functions `f(x), h(x), g(x)` as in
`h(x) = d/(dx) f(x)`
`f(x) = int_a^x g(x)dx`

    →  First part states that `h(x) = g(x)`; that is `d/(dx) int_a^x g(x) dx = g(x)`
derivative of integral of a function is the function

    →  First part implies that to find `f(x) = int_a^x g(x)dx`, find the inverse relationship `d/(dx) f(x) = g(x)` and thus, anti-derivative is defined.
integral of a function is anti-derivative of the function

    →  Second part states that `int_a^b h(x) dx = f(b)-f(a)`
That is `int_a^b d/(dx) f(x) dx = f(b)-f(a)`.
definite integral of derivative of a function is the difference of function evaluated at the limits

    →  Since the second part is true for any value `x` in the interval `[a,b]`, it implies that `int_a^x d/(dx) f(x) dx = f(x) +c`.
indefinite integral of derivative of a function is the function plus an initial value

Fundamental Theorem of Calculus

plain and simple summary

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Derivative and Integral are inverse operations.

 • Derivative of integral of a function is the function.

 • Integral of derivative of a function is the function or equivalent.

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In this page, the fundamental theorem of calculus is stated and proven. Anti-derivatives are introduced as the function that when differentiated, gives the integrand.


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Starting on learning "". In this page, the fundamental theorem of calculus is stated and proven. Anti-derivatives are introduced as the function that when differentiated, gives the integrand.

This topic is going to be little mathematically involved and firmly anchors some of the concepts in integrals. To prove Fundamental Theorem of Calculus, let us first understand "mean-value-theorem" and "squeeze theorem".

Mean-Value-Theorem for Derivatives: For a continuous and differentiable function in the closed interval `[a,b]`, there exists `c` such that `a<=c<=b`
`d/(dx)f(x)|_(x=c)` `= (f(b)-f(a)) //(b-a)`
The same can be written as
`f′(c)` `= (f(b)-f(a)) //(b-a)`. differential mean value theorem The figure graphically explains the mean value theorem :

 • ` (f(b)-f(a)) //(b-a)` is illustrated by the slope of hypotenuse of red triangle. In this, numeration and denominator are the two arms of right angle.

 • `d/(dx)f(x)|_(x=c)` is illustrated by the slope of tangent given by orange line

Which of the following is correct?

  • the tangent at `c` and the hypotenuse shown between `a` and `b` have same slope
  • there is at-least one point `c` in which slope of tangent equals slope of the hypotenuse of red triangle
  • both the above

The answer is "both the above"

Mean-Value-Theorem for integrals: For a continuous and integrable function in the closed interval `[a,b]`, there exists `c` such that `a<=c<=b`
`int_a^b f(x)dx ``= f(c) xx (b-a)`
The same can be written as
`F(b)-F(a) ``= f(c) xx (b-a)`
where `F(x)` is the integral of `f(x)`, given by `F(x) = int_a^x f(x)dx`. integral mean value theorem The figure graphically explains the mean value theorem:

 • `F(b)-F(a)` is area under the curve shown in red.

 • `f(c) xx (b-a)` is the area of rectangle shown in green.
Note: There is substantial overlap between red and green area in the figure.

Which of the following is correct?

  • area under the curve in red and the area of the rectangle in green are equal
  • there is at-least one point `c` in which area of the rectangle equals the area under the curve
  • both the above

The answer is "both the above"

Does the name "mean-value-theorem" provide any clues as to what the theorem is about?

  • Yes, one of the values in an interval is the mean for the interval.
  • No, this does not have any meaning.

The answer is "Yes, one of the values in an interval is the mean for the interval."

Squeeze theorem: A value `c` is given such that `a<=c<=b`. If `a-b` tends to `0`, then `c=a=b`.

Does the name of the theorem provide any clues as to what the theorem is about?

  • Yes, When an interval `a` and `b` is squeezed to `a=b`, then a value between these two `c` is squeezed to be equal to the two values.
  • No, this does not have any meaning.

The answer is "Yes, When an interval `a` and `b` is squeezed to `a=b`, then a value between these two, `c` is squeezed to be equal to the two values."

Given a continuous function `f(x)` in a closed interval `[a,b]` and differentiable in open interval `(a,b)`.
`h(x) = d/(dx) f(x)`
`f(x) = int_a^x g(x)dx`

First part of Fundamental Theorem of Calculus states that `h(x) = g(x)`; that is `d/(dx) int_a^x g(x) dx = g(x)`

Note: Derivative of integral of a function is the function.

What does the above mean?

  • derivative is the inverse of integral
  • derivative and integral are not related

The answer is "derivative is the inverse of integral".

Given a continuous function `f(x)` in a closed interval `[a,b]` and differentiable in open interval `(a,b)`.
`h(x) = d/(dx) f(x)`
`f(x) = int_a^x g(x)dx`

First part of Fundamental Theorem of Calculus states that `h(x) = g(x)`; that is `d/(dx) int_a^x g(x) dx = g(x)`

First part also implies the abstraction "indefinite integral" or anti-derivatives.
`int g(x)dx = c+ int_0^x g(x)dx`.
Note that the left hand side is indefinite integral and right hand side involves definite integral.

In indefinite integral, `f(x) = int_a^x g(x)dx`, the function `f(x)` is derived from aggregate of change or, equivalently, by limit of summation.

In anti-derivative, `f(x) +c = int g(x)dx`, the function `f(x)` is derived from the inverse operation of derivative. That is, derive `f(x)` such that, derivative of `f(x)` is `g(x)`.

In other words, `f(x) +c = int g(x)dx` means the anti-derivative of `g(x)` is `f(x)` (even though the representation uses integral symbol). The name anti-derivatives is coined to specify this.

What is the other name for indefinite integral?

  • anti-derivatives
  • algebraic integral

The answer is "anti-derivatives"

Given a continuous function `f(x)` in a closed interval `[a,b]` and differentiable in open interval `(a,b)`.
`h(x) = d/(dx) f(x)`
`f(x) = int_a^x g(x)dx`

Second part of Fundamental Theorem of Calculus states that `int_a^b h(x) dx = f(b)-f(a)`
That is `int_a^b d/(dx) f(x) dx = f(b)-f(a)`.

Integral of derivative of a function is the function evaluated at the limits of integral.
(stated for definite integral.)

What does the above mean?

  • integral is the inverse of derivative
  • integration of a derivative is not possible

The answer is "integral is the inverse of derivative"

Given a continuous function `f(x)` in a closed interval `[a,b]` and differentiable in open interval `(a,b)`.
`h(x) = d/(dx) f(x)`
`f(x) = int_a^x g(x)dx`

Second part of Fundamental Theorem of Calculus states that `int_a^b h(x) dx = f(b)-f(a)`
That is `int_a^b d/(dx) f(x) dx = f(b)-f(a)`.

Since the second part is true for any value `x` in the interval `[a,b]`, it can be given as `int_a^x d/(dx) f(x) dx = f(x) +c`.

Integral of derivative of a function is the function with a constant of integration.
(stated for indefinite integral.)

What does the above mean?

  • integral is the inverse of derivative
  • integration of a derivative is not possible

The answer is "integral is the inverse of derivative"

Proof for First part of Fundamental Theorem: Given a continuous function `f(x)` in a closed interval `[a,b]` and differentiable in open interval `(a,b)`.
`h(x) = d/(dx) f(x)`
`f(x) = int_a^x g(x)dx`

First part of Fundamental Theorem of Calculus states that `h(x) = g(x)`; that is `d/(dx) int_a^x g(x) dx = g(x)`

`h(x)`

`=d/(dx) f(x)`

substituting definition of differentiation
`= lim_(delta->0) (f(x+delta) - f(x)) / delta`

substituting definition of `f(x)`
`= lim_(delta->0) (int_a^(x+delta) g(x)dx ``- int_a^(x) g(x)dx) // delta`

`= lim_(delta->0) (int_a^x g(x)dx ``+ int_x^(x+delta) g(x)dx ``- int_a^(x) g(x)dx) // delta`

`= lim_(delta->0) (int_x^(x+delta) ``g(x)dx) // delta`

by mean-value-theorem `int_x^(x+delta) g(x)dx = g(c)delta` where `x<=c<=x+delta`
`= lim_(delta->0) (g(c)delta) // delta`

canceling non-zero value `delta`
`= lim_(delta->0) g(c)`

by squeeze value theorem: as `delta` tends to zero, `x<=c<=x+delta` will imply `c=x`.
`= g(x)`

What does the above prove?

  • `d/(dx) int_a^x g(x) dx = g(x)`
  • `int_a^x d/(dx) f(x) dx = f(x)+c`

The answer is "`d/(dx) int_a^x g(x) dx = g(x)`"

Proof for Second part of Fundamental Theorem: Given a continuous function `f(x)` in a closed interval `[a,b]` and differentiable in open interval `(a,b)`.
`h(x) = d/(dx) f(x)`
`f(x) = int_a^x g(x)dx`

Second part of Fundamental Theorem of Calculus states that `int_a^b h(x) dx = f(b)-f(a)`
That is `int_a^b d/(dx) f(x) dx = f(b)-f(a)`.

To simplify the statements, take `a=0`
`f(b)-f(0)`

partition the interval into `n` segments and add each of the segment
`= sum_(i=1)^n (f((b i)/n)-f((b(i-1))/n)`

by mean value theorem, `f((b i)/n)- f((b (i-1))/n) ``= b/n xx d/(dx) f(x)|_(x=c) ` where `(b i)/n <= c <=(b (i-1))/n`
`= sum_(i=1)^n b/n d/(dx) f(x)`

`h(c) = d/(dx) f(x)|_(x=c)`
`= sum_(i=1)^n h(c) b/n`

applying limit `n->oo` on both sides, the LHS is a constant so it remains unaffected and not shown
`= lim_(n->oo) sum_(i=1)^n h(c) b/n`

by squeeze value theorem as `n->oo`, the interval `(b i)/n - (b (i-1))/n = 0` and so `c =(b i)/n`
`= lim_(n->oo) sum_(i=1)^n h((bi)/n) b/n`

by definition of integral
`= int_0^b h(x)dx `

What does the above prove?

  • `d/(dx) int_a^x g(x) dx = g(x)`
  • `int_a^x d/(dx) f(x) dx = f(x)+c`

The answer is "`int_a^x d/(dx) f(x) dx = f(x)+c`"

The significance of Fundamental Theorem of Calculus:
The relationship between derivative and integral is understood as inverse operations.

    Summary of what we have learned:

     •  Differentiation, by the first principles, is the instantaneous-rate-of-change.

     •  Integration, by the first principles, is the continuous-aggregate-of-change.

     •  Integration is addition of change to an initial value.

     •  Indefinite integral is defined as a function of variable, where integration is carried out between `0` and variable `x`, along with an initial constant.

     •  Definite integral is defined as the quantity added for a change in interval `a` and `b`. This results in a numerical value.

     •  First part of fundamental theorem of calculus states that derivative of integral of a function is the function itself.

     •  Second part of fundamental theorem of calculus states that integral of derivative of a function is the function or an equivalent numerical result.

    comprehensive information for quick review

    Jogger

    comprehensive information for quick review

    Jogger

    dummy

    Fundamental Theorem of Calculus: Given a continuous function `f(x)` in a closed interval `[a,b]` and differentiable in open interval `(a,b)`.
    `h(x) = d/(dx) f(x)`
    `f(x) = int_a^x g(x)dx`

    First part states that `h(x) = g(x)`; that is `d/(dx) int_a^x g(x) dx = g(x)`

    First part implies that to find `f(x) = int_a^x g(x)dx`, find the inverse relationship `d/(dx) f(x) = g(x)` and thus, anti-derivative is defined.

    Second part states that `int_a^b h(x) dx = f(b)-f(a)`
    That is `int_a^b d/(dx) f(x) dx = f(b)-f(a)`.

    Since the second part is true for any value `x` in the interval `[a,b]`, it implies that `int_a^x d/(dx) f(x) dx = f(x) +c`.



               

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    This topic is going to be little mathematically involved and firmly anchors some of the concepts in integrals. To prove Fundamental Theorem of Calculus, let us first understand "mean-value-theorem" and "squeeze theorem".
    mean value theorem for derivatives: For a continuous and differential function in the closed interval a to b, there exists c such that, a, less than or equal, c, less than or equal b, d by d x, of f of x, at x = c = f of b minus f of a divided by b minus a. The same can be written as f prime c . The figure graphically explains the mean value theorem. Which of the following is correct?
    slope;tangent;hypotenuse
    the tangent at c and the hypotenuse shown between a and b have same slope
    there;which;red;triangle
    there is at-least one point c in which slope of tangent equals slope of the hypotenuse of red triangle
    both;above
    both the above
    The answer is "both the above"
    mean value theorem for integrals: for a continuous and integrable function in the closed interval a to b. There exists c such that, a, less than or equal, c, less than or equal b, integral a to b f of x d x, = f of c multiplied b minus a. The same can be written as capital F of b minus capital f of a equals f of c multiplied b minus a. Where capital f of x is the integral of f of x, given by capital f of x, = integral a to x f of x dx. The figure graphically explains the mean value theorem. which of the following is correct.
    area;curve;red;green
    area under the curve in red and the area of the rectangle in green are equal
    at;least;1;point
    there is at-least one point c in which area of the rectangle equals the area under the curve
    both;above
    both the above
    The answer is "both the above"
    Does the name "mean-value-theorem" provide any clues as to what the theorem is about?
    yes;s;1;values;interval
    Yes, one of the values in an interval is the mean for the interval.
    no;not;any
    No, this does not have any meaning.
    The answer is "Yes, one of the values in an interval is the mean for the interval."
    Squeeze theorem: A value c is given such that, a, less than or equal, c, less than or equal, b. If a minus b tends to 0, then c equals a equals b. Does the name of the theorem provide ant clues as to what the theorem is about.
    1
    2
    The answer is "Yes, When an interval a and b is squeezed to a=b , then a value between these two, c is squeezed to be equal to the two values."
    Given a continuous function f of x in a closed interval a to b, and differentiable in the open interval a to b. ;; h of x = derivative of f of x ;; and f of x = integral of g of x ;; First part of fundamental theorem of calculus states that h of x equals g of x. That is derivative of integral of g of x = g of x. Note: Derivative of integral of a function is the function. ;; What does the above mean.
    inverse
    derivative is the inverse of integral
    related
    derivative and integral are not related
    The answer is "derivative is the inverse of integral".
    Repeating the first part of fundamental theorem of calculus. First part also implies the abstraction indefinite integral or anti-derivatives. integral g of x d x = c + integral 0 to x, g of x d x. Note that the left hand side is the indefinite integral and right hand side involves definite integral. ;; In indefinite integral, f of x = integral a to x, g of x d x, the function f of x is derived from aggregate of change or equivalently, by limit of summation. ;; In anti-derivative f of x + c = integral g of x d x, the function f of x is derived from the inverse operation of derivative. That is, derive f of x such that, derivative of f of x is g of x . In other words, f of x + c = integral g of x dx means the anti-derivative of g of x is f of x. The name anti-derivative is coined to specify this. What is the other name for indefinite integral.
    anti-derivatives;derivatives;anti
    anti-derivatives
    algebraic;integral
    algebraic integral
    The answer is "anti-derivatives"
    Given a continuous function f of x in a closed interval a to b, and differentiable in the open interval a to b. ;; h of x = derivative of f of x ;; and f of x = integral of g of x ;; Second part of the fundamental theorem of calculus states that integral a to b of h of x d x = f of b minus f of a. That is integral a to b, d by d x of f of x d x = f of b minus f of a. Integral of derivative of a function is the function evaluated at the limits of integral. What does the above mean.
    inverse
    integral is the inverse of derivative
    possible;not
    integration of a derivative is not possible
    The answer is "integral is the inverse of derivative"
    Repeating the second part of fundamental theorem of calculus. Since the second part is true for any value x in the interval a to b, it can be given as, integral a to x, d by d x f of x d x = f of x + c. Integral of derivative of a function is the function with a constant of integration. What does the above mean.
    inverse
    integral is the inverse of derivative
    possible;not
    integration of a derivative is not possible
    The answer is "integral is the inverse of derivative"
    Proof for First part of Fundamental Theorem is given. What does this prove?
    1
    2
    The answer is "derivative of integral is the same function"
    Proof for Second part of Fundamental Theorem is given. What does the above prove.
    1
    2
    The answer is "integral of derivative is the same function"
    The significance of Fundamental Theorem of Calculus: The relationship between derivative and integral is understood as inverse operations.
    Summary of what we have learned: ;; Differentiation, by the first principles, is the instantaneous-rate-of-change. ;; Integration, by the first principles, is the continuous-aggregate-of-change. ;; Integration is addition of change to an initial value. ;; Indefinite integral is defined as a function of variable, where integration is carried out between 0 and variable x, along with an initial constant. ;; Definite integral is defined as the quantity added for a change in interval a and b. This results in a numerical value. ;; First part of fundamental theorem of calculus states that derivative of integral of a function is the function itself. ;; Second part of fundamental theorem of calculus states that integral of derivative of a function is the function or an equivalent numerical result.
    Derivative and Integral are inverse operations. ;; Derivative of integral of a function is the function. ;; Integral of derivative of a function is the function or equivalent.
    Fundamental Theorem of Calculus: Given a continuous function f of x in a closed interval a to b, and differentiable in the open interval a to b. h of x is the derivative of f of x. f of x is the indefinite integral of g of x ;; First part state that h of x = g of x ; that is derivative of integral of a function g of x is the function g of x itself. First part implies that to find f of x = integral a to x g of x dx, find the inverse relationship d by dx of f of x = g of x and thus, anti-derivative is defined. Second part states that integral a to b h of x d x = f of b minus f of a. That is integral a to b, d by d x of f of x, dx = f of b minus f of a. Since the second part is true for any value of x in the interval a to b, it implies that integral from a to x d by d x of f of x = f of x + c.

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