Introduction To Integration

I. Integration As A Method To Find Area

Let us take a look at the curve below. How to find the exact area of the region bounded by the blue curve and the $x$-axis?

Let us start by looking at at a method which will give us approximate area of the region.

We can divide the length between the two intersection points of the $x$-axis with the curve into $n$ equal line segments of length $\Delta x$ each.

Let us say the $x$ value of the midpoint of the $i\xasuper{th}$ segment be $x_i$.

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The height of each of these rectangles will be $f(x_i)$ and the width will be $\Delta x$, which will give us the area as $f(x_i) \cdot \Delta x$

Adding the areas of all the rectangles we created will give us an approximate value of the area under the curve.

Therefore, if we divide the $x$ intercept into $n$ equal segments, the area $(A)$ under the curve can be approximated as:

$A \approx \sum\limits_{i = 1}^{n} f(x_i) \cdot \Delta x$

Try the widget in the previous page, with various values of $n$. You will observe, as you increase $n$, $\Delta x$ becomes progressively smaller, and the value of the calculated area becomes closer and closer to the actual value of the area, and the error approaches $0$.

If we divide the $x$ intercept into an infinitely large number of segments, then the width of each strip, $\Delta x$, will tend to $0$, and adding the areas of all these strips, we will get the exact value of the area.

Therefore, we can write:

$A = \lim\limits_{\Delta x \rightarrow 0}\left(\sum\limits_{i = 1}^{n} f(x_i) \cdot \Delta x\right)$

In the above expression, as $\Delta x$ tends to $0$, we can replace the summation with the integration symbol $\left( \int \right)$ to express the area as:

$A = \displaystyle{\int\limits_{x_1}^{x_n}} f(x)dx$

If we take the two intersection points of the curve with the $x$-axis as $(a, 0)$ and $(b, 0)$, then as $n \rightarrow \infty$ and $\Delta x \rightarrow 0$:

$x_1 \rightarrow a$ and $x_n \rightarrow b$.

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Therefore, we can rewrite the expression for the area as:

$A = \displaystyle{\int\limits_{a}^{b}} f(x)dx$

II. Integration As Inverse Of Differentiation

We saw earlier how to use to find the slope of a curve at some given point, and in the previous section we saw how to use integration to find the area of a given curve. In this section we will develop an intuitive understanding as to why is integration the inverse operation of differentiation. That is, if

$f'(x) = \dfrac{d}{dx} f(x)$ then,

$\displaystyle \int f'(x)dx = f(x)$

The widget in the following page shows two plots.

The curve in blue shows the function $f(x) = \sqrt{x}$

The curve in green shows the area under the previous curve, from $0$ to $x$, that is:

$A(x) = \displaystyle{\int\limits_{0}^{x} f(x)dx}$

The region in red plus the region in blue is the area under the curve, from $0$ to $x + \Delta x$, that is:

$A(x + \Delta x) = \displaystyle{\int\limits_{0}^{x+\Delta x} f(x)dx}$

The blue and red shaded region is the area under the curve $f(x)$ from point $x$ to point $x + \Delta x$, and can be represented as:

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$\displaystyle A = \int\limits_{a}^{b} f(x)dx = \int\limits_{0}^{b} f(x)dx - \int\limits_{0}^{a} f(x)dx = A(x + \Delta x) - A(x)$

where, $a = x$ and $b = x + \Delta x$

Since $A(x)$ is the plot of the area under the curve $f(x)$ from $0$ to $x$, we can directly read the value of the area covered by $f(x)$ from $0$ to $p$, by reading the corresponding value of $A(x)$ at $x = p$. Therefore, $A(a) - A(b)$ will give us the exact value of the area under $f(x)$ from point $a$ to $b$. The same area is represented by the region under $f(x)$ shaded in brown.

Try to drag the blue and the red dots, and bring them as close as possible to each other.

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You will observe that as $\Delta x$ becomes closer to $0$, $f(x) \cdot \Delta x$ becomes almost equal to $A(x + \Delta x) - A(x)$. This is true, because as we saw in the previous section, as the width of the rectangle we draw under the curve approaches $0$, the error in area estimate also tends to $0$.

Therefore, using limits, we can say that,

$\displaystyle \lim\limits_{\Delta x \rightarrow 0} A(x + \Delta x) - A(x) = \lim\limits_{\Delta x \rightarrow 0} f(x) \cdot \Delta x$

$\Rightarrow \lim\limits_{\Delta x \rightarrow 0} \dfrac{A(x + \Delta x) - A(x)}{\Delta x} = f(x)$

(Note that since the $R.H.S.$ of the above equation is independent of $\Delta x$, we removed the limit operation from that side).

The $L.H.S.$ is the definition of $\dfrac{d}{dx} A(x)$

Therefore,

$\dfrac{d}{dx} A(x) = f(x)$

Observe that we started by taking $A(x) = \displaystyle \int f(x)dx$ and we have shown that $\dfrac{d}{dx} A(x) = f(x)$, that is, differentiating $A(x)$ w.r.t $x$ we got back $f(x)$. Therefore, integration and differentiation are inverse operations of each other.

This theorem, is also known as the Fundamental Theorem Of Calculus.

III. Definite And Indefinite Integrals

All the integration examples we saw so far, were aimed at finding the area between two finite points under a curve. These are called definite integrals.

However, given a function $f(x)$, if we just wanted to find a function $F(x)$ such that $\dfrac{d}{dx} F(x) = f(x)$, how can we do that?

We know that $\dfrac{d}{dx}c = 0$ for any constant $c$

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Therefore, if we find any function $F(x)$ such that, $\dfrac{d}{dx} F(x) = f(x)$, then for all constant values of $k$,

$\dfrac{d}{dx} [F(x) \pm k] = f(x)$

or, taking the inverse operation,

$F(x) \pm k = \displaystyle \int f(x)dx$

$\Rightarrow F(x) = \displaystyle \left[ \int f(x)dx \right] \mp k$

Since $k$ is a constant, its sign does not effect the result of differentiation, we can replace $\mp k$ with $c$ and write it as:

$F(x) = \displaystyle \left[ \int f(x)dx \right] + c$

where, $c$ is called the constant of integration.

Now try the following question:

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Definite integrals are always marked with the starting $x$ value, $x_1$ and the ending $x$ value, $x_2$ like:

$\displaystyle \int\limits_{x_1}^{x_2} f(x)dx$

and if in the indefinite form we have:

$\displaystyle \int f(x)dx = F(x) + c$,

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then using the definite form we can write:

$\displaystyle \int\limits_{x_1}^{x_2} f(x)dx$

$= \left(F(x) + c \right)\biggr\rvert_{x_1}^{x_2}$

$= F(x_2) + \cancel{c} - F(x_1) - \cancel{c}$

$= F(x_2) - F(x_1)$

In definite integrals, the constant always cancels out, therefore it is OK to omit the same.

The above computations are applicable if, and only if, $f(x)$ is continuous in the range $[x_1, x_2]$