11.0 Newton-Leibnitz Rule

(A) If $\phi \left( x \right)$ and $\psi \left( x \right)$ are defined on $\left[ {a,b} \right]$ and differentiable for every $x$ and $f\left( t \right)$ is continuous, then $${d \over {dx}}\int\limits_{\phi \left( x \right)}^{\psi \left( x \right)} {f\left( t \right)dt} = f\left[ {\psi \left( x \right)} \right] \cdot \psi '\left( x \right) - f\left[ {\phi \left( x \right)} \right] \cdot \phi '\left( x \right)$$

(B) If $\phi \left( x \right)$ and $\psi \left( x \right)$ are defined on $\left[ {a,b} \right]$ and differentiable for every $x$ and $f\left( t \right)$ is continuous, then $${d \over {dx}}\int\limits_{\phi \left( x \right)}^{\psi \left( x \right)} {f\left( {x,t} \right)dt} = \int\limits_{\phi \left( x \right)}^{\psi \left( x \right)} {{\delta \over {\delta x}}f\left( {x,t} \right)dt} + f\left[ {x,\psi \left( x \right)} \right] \cdot {{d\psi \left( x \right)} \over {dx}} - {d \over {dx}}\left[ {f\left[ {x,\phi \left( x \right)} \right] \cdot \phi \left( x \right)} \right]$$

(C) Let a function $f\left( {x,\alpha } \right)$ be continuous for $x \in \left[ {a,b} \right]$ and $\alpha \in \left[ {c,d} \right]$, then for any $\alpha \in \left[ {c,d} \right]$, if $$I\left( \alpha \right) = \int\limits_a^b {f\left( {x,\alpha } \right)dx} $$ then $${{dI\left( \alpha \right)} \over {d\alpha }} = \int\limits_a^b {{\delta \over {\delta \alpha }}f\left( {x,\alpha } \right)dx} $$

This can be used to solve the integrals containing two variables.

Question 25. $I = \int\limits_0^1 {{{{{\tan }^{ - 1}}ax} \over {x\sqrt {1 - {x^2}} }}dx} $

Solution: $I$ is a function of $a$ only, so $${{dI} \over {da}} = \int\limits_0^1 {{d \over {da}}{{{{\tan }^{ - 1}}ax} \over {x\sqrt {1 - {x^2}} }}dx} $$ $${{dI} \over {da}} = \int\limits_0^1 {{1 \over {x\sqrt {1 - {x^2}} }} \cdot {1 \over {1 + {a^2}{x^2}}}xdx} $$ $${{dI} \over {da}} = \int\limits_0^1 {{1 \over {\sqrt {1 - {x^2}} }} \cdot {1 \over {1 + {a^2}{x^2}}}dx} $$ Let $x = \sin t$ $$dx = \cos tdt$$ $${{dI} \over {da}} = \int\limits_0^{{\pi \over 2}} {{1 \over {\sqrt {1 - {{\sin }^2}t} }} \cdot {1 \over {1 + {a^2}{{\sin }^2}t}}\cos tdt} $$ $${{dI} \over {da}} = \int\limits_0^{{\pi \over 2}} {{1 \over {1 + {a^2}{{\sin }^2}t}}dt} $$ $${{dI} \over {da}} = \int\limits_0^{{\pi \over 2}} {{{{{\sec }^2}t} \over {{{\sec }^2}t + {a^2}ta{n^2}t}}dt} $$ $${{dI} \over {da}} = \int\limits_0^{{\pi \over 2}} {{{{{\sec }^2}t} \over {1 + \left( {{a^2} + 1} \right)ta{n^2}t}}dt} $$ Let $\left( {\sqrt {{a^2} + 1} } \right)tant = u$ $$\left( {\sqrt {{a^2} + 1} } \right){\sec ^2}tdt = du$$ $${{dI} \over {da}} = {1 \over {\sqrt {{a^2} + 1} }}\int\limits_0^\infty {{1 \over {1 + {u^2}}}du} $$ $${{dI} \over {da}} = {1 \over {\sqrt {{a^2} + 1} }}\left[ {{{\tan }^{ - 1}}u} \right]_0^\infty $$ $${{dI} \over {da}} = {1 \over {\sqrt {{a^2} + 1} }}{\pi \over 2}$$ Now Integrate w.r.t. $a$, we get, $$I = \int {{1 \over {\sqrt {{a^2} + 1} }}{\pi \over 2}da} $$ $$I = {\pi \over 2}\ln \left| {a + \sqrt {{a^2} + 1} } \right| + C$$ From initial equation of $I$, we can conclude that $$a = 0 \Rightarrow I = 0 \Rightarrow C = 0$$ $$I = {\pi \over 2}\ln \left| {a + \sqrt {{a^2} + 1} } \right|$$

Question 26. $f\left( x \right) = \int\limits_{{e^{2x}}}^{{e^{3x}}} {{t \over {\ln t}}dt} $, find derivative of $f\left( x \right)$ w.r.t. $\ln x$ at $x = \ln 4$

Solution: Let $g\left( x \right) = \ln x$ $$g'\left( x \right) = {1 \over x}$$ By Newton-Leibnitz Rule, $$f'\left( x \right) = \left( {{{{e^{3x}}} \over {\ln {e^{3x}}}}} \right){d \over {dx}}{e^{3x}} - \left( {{{{e^{2x}}} \over {\ln {e^{2x}}}}} \right){d \over {dx}}{e^{2x}}$$ $$f'\left( x \right) = {{{e^{3x}}} \over {3x}}3{e^{3x}} - {{{e^{2x}}} \over {2x}}2{e^{2x}}$$ $$f'\left( x \right) = {{{e^{6x}} - {e^{4x}}} \over x}$$ Derivative of $f\left( x \right)$ w.r.t. $g\left( x \right)$ is given by $${{f'\left( x \right)} \over {g'\left( x \right)}} = {e^{6x}} - {e^{4x}}$$ At $x = \ln 4$, $${{f'\left( {\ln 4} \right)} \over {g'\left( {\ln 4} \right)}} = {e^{6\ln 4}} - {e^{4\ln 4}}$$ $${{f'\left( {\ln 4} \right)} \over {g'\left( {\ln 4} \right)}} = {4^6} - {4^4}$$ $$ = 256\left( {16 - 1} \right) = 3840$$