If F is Continuously Differentiable Then D dx of the Integral is Equal to the Integral of D dt

Let f(x) be a function defined in a closed interval [a , b] . Then the definite integral $ \displaystyle \int_{a}^{b}f(x)dx $ represents the algebraic sum of the areas of the region bounded by the curve y = f(x) and the x-axis between the lines x = a and x = b

All the regions lying above the x-axis have ' positive ' areas whereas those lying below the x-axis have ' negative ' areas.

Trapezoidal Rule

For finding the definite integral of a linear portion of a curve we can use the formula of area of trapizium and the rule applied is called Trapezoidal rule.

In the adjoining figure AD represent a linear curve and x = a and x = b are the limits of definite integral than

$ \displaystyle \int_{a}^{b}f(x)dx = Area ABCD $

$ = \frac{1}{2}(sum \;of \;parallel \;sides) \times height $

$ = \frac{1}{2}[f(a) + f(b)] (b – a) $

Example : Evaluate:  $ \displaystyle \int_{2}^{6}(2x + 3)dx $

Solution:Here, f(x) = 2x + 3

f(6) = 15 and f(2) = 7

$ \displaystyle \int_{2}^{6}(2x+3)dx $

$ \displaystyle = \frac{(6-2)}{2}(15+7) $

= 2 × 22 = 44.

First Fundamental Theorem Of Calculus:

If f (x) is a continuous function on [a, b], then $ \displaystyle F(x) = \int_{a}^{x}f(t)dt $ is differentiable at every point x in [a, b] and $ \displaystyle \frac{d}{dx}F(x) = \frac{d}{dx} \int_{a}^{x} f(t)dt = f(x) $ ∀ x ∈ (a, b) .
This is called the first fundamental theorem of calculus.

Existence of Anti – derivative of Continuous Functions:

If y = f (x) is continuous on [a, b], then there exists a function F (x) whose derivative on [a, b] is f i.e. every continuous function is the derivative of some of the functions.

In other words, every continuous function has an anti derivative. However, not every anti derivative, even when it exists, is expressible in closed form, in terms of elementary functions e.g.

$ \displaystyle \int e^{-x^2}dx $ , $ \displaystyle \int \frac{sinx}{x} dx $ , $\displaystyle \int \frac{cosx}{x} dx $ , $ \displaystyle \int \frac{1}{ln x} dx $

In all such cases, the anti derivative is obviously some new function which does not reduce to a combination of a finite number of elementary functions.

Second Fundamental Theorem Of Calculus :

If f(x) is a continuous function on [a, b] and F(x) is any anti derivative of f(x) on [a , b]  i.e. F'(x) = f (x) ∀ x ∈ (a, b) , then  $ \displaystyle \int_{a}^{b}f(x)dx = F(b) – F(a)$

(also called the Newton-Leibnitz formula).

The function F(x) is the integral of f(x) and a and b are the lower and the upper limits of integration.

Proof :From the first fundamental theorem

$ \displaystyle \frac{d}{dx} [\int_{a}^{x}f(t)dt – F(x)] = f(x) – F'(x) = 0 $ (As F'(x) = f(x) given)

i.e., the expression within the bracket must be constant in the interval and hence we can write

$ \displaystyle F(x) = \int_{a}^{x}f(t)dt + c $  ∀ x ∈ [a , b] , where c is some real constant.

Thus $ \displaystyle F(b) = \int_{a}^{b}f(t)dt + c $

and , $ \displaystyle F(a) = \int_{a}^{a}f(t)dt + c = 0+c = c$

Hence, $ \displaystyle \int_{a}^{b}f(t)dt = F(b)-F(a) $

The second fundamental theorem is used to calculate the value of the definite integral. A note of caution to you is that f (t) must be continuous in [a, b] or else you will have to partition it into subintervals such that f(x) is continuous in each of the subintervals.

Change Of Variables In Definite Integral :

If the functions f(x) is continuous on [a, b] and the function x = g(t) is continuously differentiable on the interval [t₁, t₂] and a = g (t₁) and b = g (t₂), then

$ \displaystyle \int_{a}^{b}f(x)dx = \int_{t_1}^{t_2}f(g(t)) g'(t)dt $

Improper integrals :

Definite integral for which a or b or both are infinite are called improper integrals. These are evaluated in the following manner:

(i) $ \displaystyle \int_{a}^{\infty}f(x)dx = \lim_{b\rightarrow + \infty} \int_{a}^{b}f(x)dx $

(ii) $ \displaystyle \int_{-\infty}^{b}f(x)dx = \lim_{a\rightarrow – \infty } \int_{a}^{b}f(x)dx $

(iii) $ \displaystyle \int_{-\infty}^{+\infty}f(x)dx = \int_{-\infty}^{c}f(x)dx + \int_{c}^{+\infty}f(x)dx $

Another type of improper integral is that in which the integrand is not defined for a point c ∈ [a, b].

In both type of improper integral we take limiting values as shown in the following example.

Solved Example : Prove that : $ \displaystyle \int_{0}^{\infty} \frac{dx}{(x+\sqrt{x^2 +1})^n} = \frac{n}{n^2 -1} \; , n > 1 $

Solution: Let $ \displaystyle I = \int_{0}^{\infty} \frac{dx}{(x+\sqrt{x^2 +1})^n} $

Put $ \displaystyle x+ \sqrt{x^2 + 1} = t $

$ \displaystyle (1+\frac{x}{\sqrt{x^2+1}})dx = dt $

$ \displaystyle \frac{t dx}{\sqrt{x^2 +1}} = dt $

$ \displaystyle dx = \sqrt{1+x^2}\frac{dt}{t} $   ….(i)

$ \displaystyle \sqrt{x^2 +1} +x = t $

$ \displaystyle \sqrt{x^2 +1} – x = \frac{1}{t} $

Adding ,

$ \displaystyle 2(\sqrt{x^2 +1} ) = t + \frac{1}{t} $

$ \displaystyle (\sqrt{x^2 +1} ) =  \frac{t^2 + 1}{2 t} $

$ \displaystyle dx = \frac{t^2 + 1}{2t^2} dt $

When x → 0, t → 1 and

when x → ∞ , t → ∞

Hence ,

$ \displaystyle I = \int_{1}^{\infty} \frac{t^2 +1}{2t^2 . t^n} dt $

$ \displaystyle I = \int_{1}^{\infty} (\frac{1}{t^n} + \frac{1}{t^{n+2}} ) dt $

$ \displaystyle = \frac{1}{2} \lim_{a\rightarrow +\infty}[(\frac{-1}{(n-1)t^{n-1}} – \frac{1}{(n+1)t^{n+1}}) ]_{1}^{a} $

$ \displaystyle = \frac{1}{2} [-0-0-(\frac{-1}{n-1}-\frac{1}{n+1})] $

$ \displaystyle = \frac{n}{n^2 -1} $

Solved Example : Find the value of $\displaystyle \int_{0}^{\infty} \frac{x log x}{(1+x^2)^2} dx $

Solution: Since function changes its nature at x = 1

$\displaystyle I = \int_{0}^{\infty} \frac{x log x}{(1+x^2)^2} dx $

$\displaystyle I = \int_{0}^{1} \frac{x log x}{(1+x^2)^2} dx + \int_{1}^{\infty} \frac{x log x}{(1+x^2)^2} dx $

I = I₁ + I₂ (say)

$\displaystyle I_2 = \int_{1}^{\infty} \frac{x log x}{(1+x^2)^2} dx $

Let x = 1/y, dx = –1/y² dy

$\displaystyle I_2 = \int_{1}^{\infty} \frac{\frac{1}{y} log \frac{1}{y}}{(1+ (\frac{1}{y^2})^2)^2} (-\frac{1}{y^2}) dy $

$\displaystyle I_2 = \int_{1}^{0} \frac{y log y}{(1 + y^2)^2} dy $

$\displaystyle I_2 = – \int_{0}^{1} \frac{x log x}{(1 + x^2)^2} dx = -I_1 $

Hence, I = 0

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