3.0 Fundamental Theorems

Let $f$ be a continuous function on the closed interval $\left[ {a,b} \right]$ and let $A\left( x \right)$ be the area function. Then $$A'\left( x \right) = f\left( x \right)$$ for all $x \in \left[ {a,b} \right]$.

It is important theorem which provide us to evaluate definite integral with the use of anti-derivative.

Let $f$ be continuous function defined on the closed interval $\left[ {a,b} \right]$ and $F$ be an anti derivative of $f$. Then $$\int\limits_a^b {f\left( x \right)dx} $$ $$ = \left[ {F\left( x\right)} \right]_a^b$$ $$ = F\left( b \right) - F\left( a \right)$$

1. Find the indefinite integral $$\int {f\left( x \right)dx} $$ Let this be ${F\left( x \right)}$.

There is no need to keep integration constant $C$ because if we consider ${F\left( x \right) + C}$ instead of ${F\left( x \right)}$, we get

$$\int\limits_a^b {f\left( x \right)dx} = \left[ {F\left( x \right) + C} \right]_a^b = \left[ {F\left( b \right) + C} \right] - \left[ {F\left( a \right) + C} \right] = F\left( b \right) - F\left( a \right)$$ Thus, the arbitrary constant disappears in evaluating the value of the definite integral.

2. Evaluate $F\left( b \right) - F\left( a \right)$ as it is the value of definite integral.

Question 1. $I = \int\limits_0^{{\pi \over 4}} {\sqrt {1 + \sin 2x} dx} $

Solution: $$I = \int\limits_0^{{\pi \over 4}} {\sqrt {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} dx} $$ $$I = \int\limits_0^{{\pi \over 4}} {\sqrt {{{\left( {\sin x + \cos x} \right)}^2}} dx} $$ $$I = \int\limits_0^{{\pi \over 4}} {\left( {\sin x + \cos x} \right)dx} $$ $$I = \left[ { - \cos x + \sin x} \right]_0^{{\pi \over 4}}$$ $$I = \left[ { - {1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}} \right] - \left[ { - 1 + 0} \right]$$ $$I = 1$$

Question 2. $I = \int\limits_0^{{\pi \over 2}} {{{\tan x} \over {1 + {m^2}{{\tan }^2}x}}dx} $

Solution: Let $m\tan x = t$ $$m{\sec ^2}xdx = dt$$ $$dx = {{dt} \over {m\left( {1 + {{\left( {t/m} \right)}^2}} \right)}} = {{mdt} \over {{m^2} + {t^2}}}$$ $$I = \int\limits_0^\infty {{1 \over {1 + {t^2}}} \cdot {t \over m} \cdot {{mdt} \over {{m^2} + {t^2}}}} $$$$I = \int\limits_0^\infty {{t \over {\left( {{t^2} + 1} \right)\left( {{t^2} + {m^2}} \right)}}dt} $$ By Partial Fraction, $${t \over {\left( {{t^2} + 1} \right)\left( {{t^2} + {m^2}} \right)}} = {{At + B} \over {\left( {{t^2} + 1} \right)}} + {{Ct + D} \over {\left( {{t^2} + {m^2}} \right)}}$$ $$t = \left( {At + B} \right)\left( {{t^2} + {m^2}} \right) + \left( {Ct + D} \right)\left( {{t^2} + 1} \right)$$ $$t = {t^3}\left( {A + C} \right) + {t^2}\left( {B + D} \right) + t\left( {A{m^2} + C} \right) + \left( {B{m^2} + D} \right)$$ By comparing, we get, $$A = {1 \over {{m^2} - 1}}$$ $$C = - {1 \over {{m^2} - 1}}$$ $$B = D = 0$$ $$I = {1 \over {{m^2} - 1}}\int\limits_0^\infty {{t \over {{t^2} + 1}} - {t \over {{t^2} + {m^2}}}dt} $$ $$I = {1 \over {2\left( {{m^2} - 1} \right)}}\left[ {\ln \left( {{t^2} + 1} \right) - \ln \left( {{t^2} + {m^2}} \right)} \right]_0^\infty $$ $$I = {1 \over {2\left( {{m^2} - 1} \right)}}\left[ {\ln \left( {{{{t^2} + 1} \over {{t^2} + {m^2}}}} \right)} \right]_0^\infty $$ $$I = {1 \over {2\left( {{m^2} - 1} \right)}}\left[ {\ln \left( {{{1 + 0} \over {1 + 0}}} \right) - \ln \left( {{{0 + 1} \over {0 + {m^2}}}} \right)} \right]$$ $$I = {1 \over {2\left( {{m^2} - 1} \right)}}\left[ {0 - \ln \left( {{1 \over {{m^2}}}} \right)} \right]$$ $$I = {{\ln {m^2}} \over {2\left( {{m^2} - 1} \right)}}$$