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.

*click on the content to continue..*

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)`. 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`. 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`.

*slide-show version coming soon*