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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). 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.

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