Definite Integration And Fundamental Theorem

GEOMETRICAL INTERPRETATION OF DEFINITE INTEGRAL

If f(x) > 0 for all x $\in$ [a, b]; then${\int }_a^{b}f(x)$is numerically equal to the area bounded by the curve y = f(x), then x-axis and the straight lines x = a and x = b i.e. ${\int }_a^{b}f(x)$

In general ${\int }_a^{b}f(x)\text{dx}$represents to algebraic sum of the figures bounded by the curve

y = f(x), the x-axis and the straight line x = a and x = b. The areas above x-axis are taken place plus sign and the areas below x-axis are taken with minus sign i.e,

i.e. ${\int }_a^{b}f(x)\text{dx}=$ area OLA - area AQM – area MRB + area BSCD

Note: ${\int }_a^{b}f(x)\text{dx}=$, represents algebraic sum of areas means, that if area of function y = f(x)

is asked between a to b.

$\Rightarrow$Area bounded = ${\int }_a^b\left|f(x)\right|dxandnotbeenrepresentedby{\int }_a^{b}f(x)$dx

e.g., If some one asks the area of y = x³ between -1 to 1.

Then y = x³ could be plotted as;

$\therefore$ Area = ${\int }_{-1}^0-x^{3}dx+{\int }_0^1{x}^{3}dx=\frac{1}{2}$

or, using above definition Area = ${\int }_{-1}^1\left|x^3\right|dx=2\underset{0}{\overset{1}{\int }}{x}^{3}dx$

$=2{\left[\frac{x^4}{4}\right]}_0^1=\frac{1}{2}$

But if, we integrate x³ between -1 to 1.

$\therefore$ ${\int }_{-1}^1{x}^{3}dx=0$ which does not represent area.

Thus, students are adviced to make difference between area and definiteIntegral.

FUNDAMENTAL THEOREM OF CALCULUS (NEWTON-LEIBNITZ FORMULA)

This theorem state that 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. $F^{\prime }$(x) = f (x) $\forall$ x $\in$ (a, b), then $\underset{a}{\overset{b}{\int }}f(x)dx=F(b)-F(a)$

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

Illustration : Evaluate ${\int }_{-2}^2\frac{dx}{4+x^2}$ directly as well as by the substitution x = 1 /t. Examine as to why the answer do not tally?

Solution: $I={\int }_{-2}^2\frac{dx}{4+x^2}$

\begin{align}  =\left[ \dfrac{1}{2}ta{{n}^{-1}}\left( \dfrac{x}{2} \right) \right]_{-2}^{2}=\dfrac{1}{2}\left[ ta{{n}^{-1}}\left( 1 \right)-ta{{n}^{-1}}(-1) \right] \\  =\,\dfrac{1}{2}\left[ \dfrac{\pi }{4}-\left( -\dfrac{\pi }{4} \right) \right]\,=\dfrac{\pi }{4}\Rightarrow \,I=\dfrac{\pi }{4} \\ \end{align}

On the other hand; if x = 1/t then,

$I={\int }_{-2}^2\frac{dx}{4+x^2}={\int }_{-1/2}^{1/2}\frac{dt}{t^2(4+1/t^2)}=-{\int }_{-1/2}^{1/2}\frac{dt}{4t^2+1}$

$=-{\left[\frac{1}{2}ta{n}^{-1}(2t)\right]}_{-1/2}^{1/2}$

$=-\frac{1}{2}{\tan }^{-1}(1)-\left(-\frac{1}{2}{\tan }^{-1}(-1)\right)=-\frac{\pi }{8}-\frac{\pi }{8}=-\frac{\pi }{4}$

$\therefore$ I = $\frac{\pi }{4}$ when x =$\frac{1}{t}$

In above two results l = - $\pi$/4 is wrong. Since the integrand $\frac{1}{4+x^2}>0$ and therefore the definite integral of this function cannot be negative.

Since x = 1/t is discontinuous at t = 0, the substitution is not valid ($\therefore$ I = $\pi$/4).

Note: It is important the substitution must be continuous in the interval of integration.