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mathsIntegral CalculusBasics of Integration

Fundamental Theorem of Calculus

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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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
  • 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
  • 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.
  • 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.
  • 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 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
  • 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
  • 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
  • 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)`
  • `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`
  • `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.

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(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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