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Thought-Process to Discover Knowledge

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Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

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mathsIntegral CalculusBasics of Integration

Integration: Graphical Meaning

In this page, graphical meaning of integration is discussed with examples.



click on the content to continue..

For now, integration of `f(x)` is considered in general to be

`int f(x)dx = c + int_0^x f(x)dx`

constant `c` is the initial value of integral at `x=0`.

The classification as definite integral and indefinite integral or anti-derivatives will be explained in due course.

Summary of integration: Given `y=f(x)` a function of variable `x`, the integral of `y` is

`int f(x) dx``=c+ lim_(n->oo) sum_(i=1)^(n) f((ix)/n)xx x/n`

Which of the following is correct about this definition of integral?

  • The integral is another function of variable `x`
  • The integral is another function of variable `x`
  • the integral is a numerical value

The answer is "The integral is another function of variable `x`".

the definite integral will be explained later.

Let us consider `y=x` and the integral `int xdx = x^2/2 + c`. plot of x and integral x The figure depicts both the functions. Orange color line is the given function `y=x` and the blue color curve is the integral `int xdx = x^2/2 + 0 `.
note: The plot is not to the scale on x and y axes.

How are these two graphs related?

  • the aggregate of change of orange line is plotted as blue curve
  • the aggregate of change of orange line is plotted as blue curve
  • the two plots are not related

The answer is "the aggregate of change of orange line is plotted as blue curve".

the aggregate of change is given as a function of `x`

Considering `y=x` and the integral `int xdx = x^2/2 +c` given in the figure.zoom of plot x and integral x The figure zooms in a small part of the plots. What do the two values at `x=a` (on the line and curve) signify?

  • `y` is the value function evaluates to at `x=a`
  • `int y` is the aggregate of change of y between `x=0` and `x=a`
  • both the above
  • both the above

The answer is "both the above". This result is known from the algebraic derivations. Let us see what this means in the given graphs.

Considering `y=x` and the integral `int xdx = x^2/2 +c` given in the figure.area of plot x and integral x The integral in first principles is given as

`int ydx`

`=lim_(n->oo) sum_(i=1)^n (ix)/n xx x/n`

What does the expression inside the summation `x/n` mean?

  • the interval between `0` and `x` is split into `n` partitions
  • the interval between `0` and `x` is split into `n` partitions
  • it does not denote anything

The answer is "the interval between `0` and `x` is split into `n` partitions". This is shown as `n` vertical bars placed on x axis.

Considering `y=x` and the integral `int xdx = x^2/2 +c` given in the figure.area of plot x and integral x The integral in first principles is given as

`int ydx`

`=lim_(n->oo) sum_(i=1)^n (ix)/n xx x/n`

keeping aside the limit `lim_(n->oo)`, and considering the sum with an assumed value for `n`
What does the expression inside the summation `f((ix)/n) xx x/n` mean?

  • the value of function at each of the `n` partitions is multiplied by the partition width
  • the value of function at each of the `n` partitions is multiplied by the partition width
  • it does not denote anything

The answer is "the value of function at each of the `n` partitions is multiplied by the partition width". This is shown by the height of the vertical bars of the `n` partitions on the x axis and these bars reach the line represented by the given function.

Considering `y=x` and the integral `int xdx = x^2/2 +c` given in the figure.area of plot x and integral x The integral in first principles is given as

`int ydx`

`=lim_(n->oo) sum_(i=1)^n (ix)/n xx x/n`

The sum is illustrated in the figure as vertical bars. The `n` is chosen to be `15`.
Set aside the limit `lim_(n->oo)`, and consider the sum with `n=15`
Geometrically, what does the integral given by the sum depict?

  • area under the given function `y=x` between `0` and `a`
  • area under the given function `y=x` between `0` and `a`
  • the integral does not depict anything geometrical

The answer is "area of under the given function `y=x` between `0` and `a`".

Consider a different function `y=x^2` and the integral of the function `int x^2 dx = x^3/3 +c`.integral of x square x^2 The orange color curve is the given function `y=x^2` and the blue color curve is the integral `x^3/3`
assuming the constant of integration `c=0`

How are these two plots related?

  • the aggregate of change of orange line is plotted as blue curve
  • the aggregate of change of orange line is plotted as blue curve
  • the two plots are not related

The answer is "the aggregate of change of orange line is plotted as blue curve".

the aggregate of change is a function of `x`

Considering `y=x^2` and `int y dx = x^3/3 +c` given in the figure. integral of x squared zoomed The figure zooms in a small part of the plots. What do the two values at `x=a` signify?

  • `y` is the value function evaluates to at `x=a`
  • `int y` is the aggregate of change of `y` between `x=0` and `x=a`
  • both the above
  • both the above

The answer is "both the above". This result is known from the algebraic derivations. Let us see what this means in the given curve.

Considering `y=x^2` and `int y dx = x^3/3 +c` given in the figure.zoom in area under x squared The integral in first principles is given as

`int ydx`

`=lim_(n->oo) sum_(i=1)^n ((ix)/n)^2 xx x/n`

The sum is illustrated in the figure as vertical bars. The `n` is chosen to be `10`.
Setting aside the limit `lim_(n->oo)`, and considering the sum with `n=10`

Geometrically, what does the vertical bars given by the sum depict?

  • area under curve
  • area under curve
  • the integral does not depict anything geometrical

The answer is " area under curve ".

Considering `y=x^2` and `int y dx = x^3/3 +c` given in the figure.zoom in area under x squared The integral in first principles is given as

`int ydx`

`=lim_(n->oo) sum_(i=1)^n ((ix)/n)^2 xx x/n`

The sum is illustrated in the figure as vertical bars. The `n` is chosen to be `10`.
Set aside the limit `lim_(n->oo)`, and consider the sum with `n=10`

Are the following two equal?
the area under curve `y=x^2` and the sum given by bars.

  • no, the two are not equal
  • no, the two are not equal
  • yes, the two are equal

The answer is "No, the two are not equal". The difference between area under the curve and the sum with `n=10` is highlighted in green.

Considering `y=x^2` and `int y dx = x^3/3 +c` given in the figure.zoom in area under x squared The integral in first principles is given as

`int ydx`

`=lim_(n->oo) sum_(i=1)^n ((ix)/n)^2 xx x/n`

The `n` is chosen to be `20` and the vertical bars depict the sum.
Set aside the limit `lim_(n->oo)`, and consider the sum with `n=20`

What is happening to the difference between the area under curve `y=x^2` and the sum given by the bars?

  • reducing
  • reducing
  • increasing

The answer is "reducing". The difference between area under the curve and the sum with `n=20` is lesser than the same difference with `n=10`.

Considering `y=x^2` and `int y dx = x^3/3 +c` given in the figure.zoom in area under x squared The integral in first principles is given as

`int ydx`

`=lim_(n->oo) sum_(i=1)^n ((ix)/n)^2 xx x/n`

As the limit `n` tending to infinity is applied, what will be the area under the curve and the sum given by vertical bars?

  • identical
  • identical
  • will have some non-zero difference

The answer is "identical". Note that the function is assumed to be continuous and integrable. These will be explained later in detail.

`int_0^x y dx |_(x=a)` is the area under the curve between y axis and `x=a`.plot of x squared with area. Note 1: The definition of the integral is taken to be `int_0^x f(x)dx`

Note 2: The constant of integration (c) is not explained yet.

Note 3: In this definition, the area under the curve is signed area, that is it can have both positive and negative values.

These are explained in the subsequent questions with examples.

Considering `y=x^2` in orange.zoom in area under x squared The integral is the aggregate of change added to the initial value of a quantity. That is, the quantity (which is the result of summation carried out in integration) might have a non-zero initial value.

 •  The curve in blue shows the `int x^2 dx` with an initial value `y_0=0`.

 •  The curve in red shows the `int x^2 dx` with an initial value `y_1` : a positive value.

Could you find the initial value of the curve in green?

  • positive value
  • 0
  • 0

The answer is "negative value".

We have understood the initial value as constant in abstraction. Does that have any physical significance in applications?sine x integration To explore this, the function `y=sin x` and integral `int sinx dx = -cos x + c` is considered.
The result of integration will be proven later.

Given `y=sin x ` and integral of `y`, the application scenario is the velocity and displacement of a ball attached to an oscillating spring.oscillating spring initial value The position of rest is shown as position B.
The velocity `v = sin t` is plotted in orange.

The spring is pulled from position of rest B to an initial position A and released.

The displacement is measured from right to left. And it can be measured from one of the following start positions.

 • from position A, in which case, the constant is `1`.

 • from position B, in which case, the constant is `0`.

 • from position C, in which case, the constant is `-1`.

The initial value is calculated by equating the derived value to the observed value. From position A:

 • the derived value is `-cost + c` when time `t=0`

 • Observed value is `0` when time `t-0`

Equating these two `-cos0+c = 0` results in `c = 1`.

Students may work out the same for position B and position C. What are the values of constants for position B and Position C?

  • `0` and `-1`
  • `0` and `-1`
  • `1` and `-1`

The answer is "`0` and `-1`".

Consider `y` shown as orange curve and the `int y dx` shown as blue curve.
The curves are `sinx` and its integral. This will be explained in standard results of integrals.integral sin x Between `x=0` and `x=pi`, the function `y` has positive value. The area under the orange curve is increasing between `x=0` and `x=pi`. Observe the blue curve that depicts the area as a function of `x`.

What is the value of integral (shown in blue) between `x=0` and `x=pi`?

  • increasing
  • increasing
  • constant

The answer is "increasing"

Consider `y` shown as orange curve and the `int y dx` shown as blue curve.

integral sin x Between `x=0` and `x=pi`, the function `y` has positive value and the value of integral is increasing.

Between `x=pi` and `x=2pi`, the function `y` has negative value. The area under the curve is negative between `x=pi` and `x=2pi`. The negative area is added to the integral value at `x=pi`.

What is the value of integral between `x=pi` and `x=2pi`?

  • decreasing
  • decreasing
  • constant

The answer is "decreasing"

Consider `y=x` shown as orange curve and the `int y dx = x^2/2+c` shown as blue curve. For this example, `c=0`.

integral x and x squared On the negative x-axis, the area under the curve is negative. But, the integral value (given by blue curve) is positive. How is this explained?

  • the graph is not correct in negative x-axis
  • the integration is defined from `x=0` and area is negative when traversed in negative direction
  • the integration is defined from `x=0` and area is negative when traversed in negative direction

The answer is "the integration is defined from `x=0`"

`int f(x)dx = int_0^(color(coral)(x)) f(x)dx +c`
In negative x axis, the value of upper limit `color(coral)(x)` is negative. The summation is in the reverse direction from `0` to negative value of `x`.

Integral summation is defined for integration starting from y-axis, traversing to the right. But, in the negative x axis, integration traverses to the left, which is negative.

Negative area multiplied negative traversal makes the value of integration positive.

Consider `y=x^2` shown as orange curve and the `int y dx = x^3/3+c` shown as blue curve. For this example, `c=0`.

integral x and x squared On the negative x-axis, the area under the curve is positive. But, the integral values (given by blue curve) is negative. How is this explained?

  • the graph is not correct in negative x-axis
  • the integration is defined from `x=0` and that makes the positive area multiplied by negative
  • the integration is defined from `x=0` and that makes the positive area multiplied by negative

The answer is "the integration is defined from `x=0` and that makes the positive area multiplied by negative"

`int f(x)dx = int_0^(color(coral)(x)) f(x)dx +c`
In negative x axis, the summation is in the reverse direction from `0` to negative value of `color(coral)x`.

Integral summation is defined for integration starting from y-axis, traversing to the right. But, in the negative x axis, integration traverses to the left, which is negative.

Positive area multiplied negative traversal makes the value of integration negative.

We have understood the positive and negative areas under the curve in abstraction. Does that have any physical significance in applications?sine x integration To explore this, the function `y=sin x` and integral `int sinx dx = -cos x+c` is considered.

Given `y=sin x ` and integral of `y`, the application scenario is the velocity and displacement of a ball attached to an oscillating spring.oscillating spring The position of rest is shown as position B.
The velocity `v = sin t` is plotted in orange.
The displacement `s = int v dt = -cosx + 1` is plotted in blue.

 •  The spring is pulled from position of rest B to an initial position A and released.

In this position A, what is the velocity?

  • maxima and minima
  • `0`
  • `0`

The answer is "`0`". The ball is released and so its initial velocity is `0`.

considering the oscillating spring.oscillating spring

 •  At the initial position A, the velocity is `0`.

 •  The displacement is measured from that initial position A and thus is `0`.

 •  As the pulled spring is released, it gains velocity. As it reaches the position of rest B, it has gained maximum velocity. That is shown as the point B and orange curve reaches maxima.

 •  At the position of rest B, the displacement (integral of the velocity) is a positive value.

In an oscillating spring, at the position of rest, what will be the velocity?

  • maxima and minima
  • maxima and minima
  • `0`

The answer is "maxima and minima".

considering the oscillating spring.oscillating spring The ball has reached from position A to Position B with velocity reaching maxima.

 •  Because of inertia, the ball continues in the same direction, but the spring exerts force in opposite direction, as the spring is being compressed beyond position B. Thus, the velocity reduces.

 •  The displacement continues to increase.

 •  As the reducing velocity reaches `0`, the spring reaches the maximum displacement. This is shown as the point C and blue curve reaches maxima.

In an oscillating spring, at the position C, what will be the velocity?

  • maxima and minima
  • `0`
  • `0`

The answer is "`0`".

considering the oscillating spring. oscillating spring The ball was released from position A, has crossed position B, and has reached position C.

 •  The spring is compressed at position C and it exerts force on the ball. This results in velocity increasing in negative direction.

 •  The displacement reduces from the position C.

This explains, what is meant by "negative- area" under the curve.

What will be the displacement after reaching the position C?

  • decreases
  • decreases
  • increases

The answer is "decreases". This is shown by the blue curve decreasing beyond. In this interval, the velocity is in the negative direction and is shown by the orange curve.

Graphical meaning of Integrals is signed area under the curve.

Graphical meaning of Integrals:

 • definite integral represents the signed area under the curve.

 • indefinite integral can be interpreted as the signed area under the curve from `0` to `x` added to initial constant `c`

 • Area is positive if it is above the x axis and negative if it is below the x axis.

 • Traversal is positive if it is in positive x direction and negative if it is in negative x direction.

 • The sign of the integral is the product of the sign of area and sign of the traversal.

 • The constant of integration is the initial value of the quantity represented by the integral.

Solved Exercise Problem:

Given `y=x^2`, What is the length of the curve from `x=0` as a function of `x`?

  • this is not an integration problem, as integration is area under the curve
  • this can be solved using integration
  • this can be solved using integration

The answer is 'it can be solved using integration'. It is explained in the subsequent pages.

After studying that the graphical meaning of the integration is area under the curve, using integration to find the length of the curve might be confusing.

If `y=x^2` is integrated, then it is the area under the given curve.
But the question is to find length of curve. Formulate the length of the curve as given below.

 • For a small change `dx`, the change in `y` is `dy`.
 • For a small change, `dx`, and `dy` forms a right-triangle with hypotenuse as arc length
 • Small change in Arc length `= sqrt(dx^2+dy^2)`.

Aggregate of arc length is integration of small change in arc length.

Arc length `s`

` = int_0^x sqrt(dx^2 + dy^2)`

` = int_0^x sqrt(dx^2 + (2xdx)^2)`

` = int_0^x sqrt(1+ 4x^2) dx`

Integrating this expression is given in due course. The point is that instead of defining and remembering integration as area under the curve, remember that integration is aggregate of change. Area under the curve is one form, that is, when the given function is directly integrated.

Note: After the formulation for arc length is derived as `int_0^x sqrt(1+4x^2)dx`, this integration can be considered as area under the curve for the function `sqrt(1+4x^2)`. And this area is equivalently the arc length for given `y=x^2`. This is little involved to understand, hope you could.

                            
slide-show version coming soon