7.1 Proof of Properties

7.1 Proof of Properties

$$\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( t \right)dt} $$

Put $x = t$ in any integral

Question 10. $I = \int\limits_0^{{\pi \over 2}} {\sin xdx} $

Solution: Let $x = t$ in integral $$ \Rightarrow dx = dt$$

Integral I become $$I = \int\limits_0^{{\pi \over 2}} {\sin xdx} $$ $$I = \left[ { - {\mathop{\rm cost}\nolimits} } \right]_0^{{\pi \over 2}}$$ $$I = \left[ { - \cos {\pi \over 2}} \right] - \left[ { - \cos 0} \right]$$ $$I = 0 + 1 = 1$$

$$\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx} $$

Let $F$ be anti derivative of $f$. Then, by the second fundamental theorem of calculus, we have

$$\int\limits_a^b {f\left( x \right)dx} = \left[ {F\left( b \right) - F\left( a \right)} \right]$$

$$\int\limits_a^b {f\left( x \right)dx} = - \left[ {F\left( a \right) - F\left( b \right)} \right]$$

$$\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( x \right)dx} $$

Here, we observe that, if $a = b$, then $$\int\limits_a^a {f\left( x \right)dx} = 0$$

$$\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^c {f\left( x \right)dx} + \int\limits_c^b {f\left( x \right)dx} $$

Let $F$ be an derivative of $f$. Then

$$\begin{equation} \begin{aligned} \int\limits_a^b {f\left( x \right)dx} = \left[ {F\left( b \right) - F\left( a \right)} \right]\quad ...(i) \\ \int\limits_a^c {f\left( x \right)dx} = \left[ {F\left( c \right) - F\left( a \right)} \right]\quad ...(ii) \\ \int\limits_c^b {f\left( x \right)dx} = \left[ {F\left( b \right) - F\left( c \right)} \right]\quad ...(iii) \\\end{aligned} \end{equation} $$

Adding equation $(ii)$ and $(iii)$ we get,

$$\int\limits_a^c {f\left( x \right)dx} + \int\limits_c^b {f\left( x \right)dx} = \left[ {F\left( b \right) - F\left( a \right)} \right] = \int\limits_a^b {f\left( x \right)dx} $$

Question 11. $I = \int\limits_{{e^{ - 1}}}^{{e^2}} {\left| {{{\ln x} \over x}} \right|dx} $

Solution: For $x \in \left[ {{e^{ - 1}},1} \right]$, $$\left| {{{\ln x} \over x}} \right| = - {{\ln x} \over x}$$ For $x \in \left[ {1,{e^2}} \right]$, $$\left| {{{\ln x} \over x}} \right| = {{\ln x} \over x}$$ $$I = \int\limits_{{e^{ - 1}}}^1 {\left| {{{\ln x} \over x}} \right|dx} + \int\limits_1^{{e^2}} {\left| {{{\ln x} \over x}} \right|dx} $$ $$I = - \int\limits_{{e^{ - 1}}}^1 {{{\ln x} \over x}dx} + \int\limits_1^{{e^2}} {{{\ln x} \over x}dx} $$ Let ${I_0} = \int {{{\ln x} \over x}dx} $ $${I_0} = \ln x\int {{1 \over x}dx} - \int {\left( {{1 \over x} \cdot \int {{1 \over x}dx} } \right)dx} $$ $${I_0} = {\left( {\ln x} \right)^2} - {I_0}$$ $${I_0} = {{{{\left( {\ln x} \right)}^2}} \over 2}$$ Using this in $I$, we get, $$I = - \left[ {{{{{\left( {\ln x} \right)}^2}} \over 2}} \right]_{{e^{ - 1}}}^1 + \left[ {{{{{\left( {\ln x} \right)}^2}} \over 2}} \right]_1^{{e^2}}$$ $$I = - 0 + {1 \over 2} + 2 - 0$$ $$I = {5 \over 2}$$

$$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} } $$

Let $x = a + b - t$ $$ \Rightarrow dx = - dt$$ $$x = a \Rightarrow t = b$$ $$x = b \Rightarrow t = a$$ Integral become $$\int\limits_a^b {f\left( x \right)dx} = - \int\limits_b^a {f\left( {a + b - t} \right)dt} $$ Using Property $2$, $$\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - t} \right)dt} $$ Using Property $1$, $$\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$

$$\int\limits_0^a {f\left( x \right)dx} = \int\limits_0^a {f\left( {a - x} \right)dx} $$ Let $x = a - t$ $$ \Rightarrow dx = - dt$$ $$x = 0 \Rightarrow t = a$$ $$x = a \Rightarrow t = 0$$ Integral become $$\int\limits_0^a {f\left( x \right)dx} = - \int\limits_a^0 {f\left( {a - t} \right)dt} $$ Using Property 2, $$\int\limits_0^a {f\left( x \right)dx} = \int\limits_0^a {f\left( {a - t} \right)dt} $$ Using Property 1, $$\int\limits_0^a {f\left( x \right)dx} = \int\limits_0^a {f\left( {a - x} \right)dx} $$

Question 12. $I = \int\limits_0^{{\pi \over 4}} {\ln \left( {1 + \tan x} \right)dx} $

Solution: Let, $$I = \int\limits_0^{\frac{\pi }{4}} {\ln \left( {1 + \tan x} \right)dx} \quad ...(i)$$

Using Property 5, $$I = \int\limits_0^{{\pi \over 4}} {\ln \left( {1 + \tan \left( {{\pi \over 4} - x} \right)} \right)dx} $$ $$I = \int\limits_0^{{\pi \over 4}} {\ln \left( {1 + {{\tan {\pi \over 4} - \tan x} \over {1 + \tan {\pi \over 4}\tan x}}} \right)dx} $$ $$I = \int\limits_0^{{\pi \over 4}} {\ln \left( {1 + {{1 - \tan x} \over {1 + \tan x}}} \right)dx} $$ $$I = \int\limits_0^{{\pi \over 4}} {\ln \left( {{{1 + \tan x + 1 - \tan x} \over {1 + \tan x}}} \right)dx} $$ $$I = \int\limits_0^{\frac{\pi }{4}} {\ln \left( {\frac{2}{{1 + \tan x}}} \right)dx} \quad ...(ii)$$

Now adding both equation $(i)$ and $(ii)$ we get, $$2I = \int\limits_0^{{\pi \over 4}} {\ln \left( {1 + \tan x} \right) + \ln \left( {{2 \over {1 + \tan x}}} \right)dx} $$ $$2I = \int\limits_0^{{\pi \over 4}} {\ln 2dx} $$ $$I = {{\ln 2} \over 2}\left[ x \right]_0^{{\pi \over 4}}$$ $$I = {\pi \over 8}\ln 2$$

Question 13. $I = \int\limits_0^{{\pi \over 2}} {{{{{\tan }^7}x} \over {{{\tan }^7}x + {{\cot }^7}x}}dx} $

Solution: By using Property, $$I = \int\limits_0^{{\pi \over 2}} {{{{{\tan }^7}\left( {{\pi \over 2} - x} \right)} \over {{{\tan }^7}\left( {{\pi \over 2} - x} \right) + {{\cot }^7}\left( {{\pi \over 2} - x} \right)}}dx} $$ $$I = \int\limits_0^{{\pi \over 2}} {{{{{\cot }^7}x} \over {{{\tan }^7}x + {{\cot }^7}x}}dx} $$ Adding the both integrals, we get, $$2I = \int\limits_0^{{\pi \over 2}} {{{ta{n^7}x + {{\cot }^7}x} \over {{{\tan }^7}x + {{\cot }^7}x}}dx} $$ $$2I = \int\limits_0^{{\pi \over 2}} {1dx} $$ $$2I = \left[ x \right]_0^{{\pi \over 2}} = \left[ {{\pi \over 2} - 0} \right]$$ $$I = {\pi \over 4}$$

$$\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( {2a - x} \right)dx} $$

Using Property $3$, $$\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_a^{2a} {f\left( x \right)dx} $$ In the second integral, put $x = 2a - t$ $$ \Rightarrow dx = - dt$$ $$x = a \Rightarrow t = a$$ $$x = 2a \Rightarrow t = 0$$ Integral become $$\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} - \int\limits_a^0 {f\left( {2a - t} \right)dt} $$ Using Property 2, $$\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( {2a - t} \right)dt} $$ Using Property 1, $$\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( {2a - x} \right)dx} $$

Using Property $6$, $$\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( {2a - x} \right)dx} $$ In the second integral,

For $f\left( {2a - x} \right) = f\left( x \right)$ $$\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( x \right)dx} $$ $$\int\limits_0^{2a} {f\left( x \right)dx} = 2\int\limits_0^a {f\left( x \right)dx} $$ For $f\left( {2a - x} \right) = - f\left( x \right)$ $$\int\limits_0^{2a} {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} - \int\limits_0^a {f\left( x \right)dx} = 0$$

Question 14. $I = \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{1 \over {1 + \cos x}}dx} $

Solution: By using the Property, $$I = \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{1 \over {1 + \cos \left( {{{3\pi } \over 4} + {\pi \over 4} - x} \right)}}dx} $$ $$I = \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{1 \over {1 - \cos x}}dx} $$ $$2I = \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{1 \over {1 + \cos x}} + {1 \over {1 - \cos x}}dx} $$ $$2I = \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{2 \over {1 - {{\cos }^2}x}}dx} $$ $$I = \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{{\mathop{\rm cosec}\nolimits} }^2}xdx} $$ $$I = \left[ { - \cot x} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$ $$I = 1 + 1 = 2$$

Using Property $3$, $$\int\limits_{ - a}^a {f\left( x \right)dx} = \int\limits_{ - a}^0 {f\left( x \right)dx} + \int\limits_0^a {f\left( x \right)dx} $$ In first integral, Let $x = - t$

$$ \Rightarrow dx = - dt$$ $$x = - a \Rightarrow t = a$$ $$x = 0 \Rightarrow t = 0$$ Integral become $$\int\limits_{ - a}^a {f\left( x \right)dx} = - \int\limits_a^0 {f\left( { - t} \right)dt} + \int\limits_0^a {f\left( x \right)dx} $$ Using Property 1 and 2, $$\int\limits_{ - a}^a {f\left( x \right)dx} = \int\limits_0^a {f\left( { - x} \right)dx} + \int\limits_0^a {f\left( x \right)dx} $$ For $f\left( { - x} \right) = f\left( x \right)$ $$\int\limits_{ - a}^a {f\left( x \right)dx} = \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( x \right)dx} $$ $$\int\limits_{ - a}^a {f\left( x \right)dx} = 2\int\limits_0^a {f\left( x \right)dx} $$ For $f\left( { - x} \right) = - f\left( x \right)$ $$\int\limits_{ - a}^a {f\left( x \right)dx} = - \int\limits_0^a {f\left( x \right)dx} + \int\limits_0^a {f\left( x \right)dx} = 0$$

Question 15. $I = \int\limits_{ - \pi }^\pi {{{\sin }^3}x{{\cos }^2}xdx} $

Solution: Let $f\left( x \right) = {\sin ^3}x{\cos ^2}x$ $$f\left( { - x} \right) = {\sin ^3}\left( { - x} \right){\cos ^2}\left( { - x} \right) = - f\left( x \right)$$ So, function $f\left( x \right)$ is an odd function.

Integral become $$I = 0$$

For a function $f$ having time period $T$, net sum of area under the curve for each time period is same.

Question 16. $I = \int\limits_0^{28\pi } {{1 \over {1 + {e^{\sin x}}}}dx} $

Solution: Here $$f\left( x \right) = {1 \over {1 + {e^{\sin x}}}}$$ whose Time Period is $T = 2\pi $ $$I = 14\int\limits_0^{2\pi } {{1 \over {1 + {e^{\sin x}}}}dx} $$ Using Property 7, we get $$I = 14\int\limits_0^\pi {{1 \over {1 + {e^{\sin x}}}} + {1 \over {1 + {e^{\sin \left( {2\pi - x} \right)}}}}dx} $$ $$I = 14\int\limits_0^\pi {{1 \over {1 + {e^{\sin x}}}} + {1 \over {1 + {e^{ - {\mathop{\rm sinx}\nolimits} }}}}dx} $$ $$I = 14\int\limits_0^\pi {{1 \over {1 + {e^{\sin x}}}} + {{{e^{\sin x}}} \over {1 + {e^{{\mathop{\rm sinx}\nolimits} }}}}dx} $$ $$I = 14\int\limits_0^\pi {{{1 + {e^{\sin x}}} \over {1 + {e^{\sin x}}}}dx} $$ $$I = 14\left[ x \right]_0^\pi $$ $$I = 14\pi $$

Question 17. Let ${\pi \over 2} < v < \pi $ and $n \in N$ then prove that $$\int\limits_0^{n\pi + v} {\left| {\cos x} \right|dx} = 2n + 2 - \sin v$$

Solution: Let $I = \int\limits_0^{n\pi + v} {\left| {\cos x} \right|dx} $

Using Property 3, $$I = \int\limits_0^{n\pi } {\left| {\cos x} \right|dx} + \int\limits_{n\pi }^{n\pi + v} {\left| {\cos x} \right|dx} $$ $f\left( x \right) = \left| {\cos x} \right|$ is of Time Period $\pi $ so by Property 9, $$I = n\int\limits_0^\pi {\left| {\cos x} \right|dx} + \int\limits_0^v {\left| {\cos x} \right|dx} $$ $$I = n\left[ {\int\limits_0^{{\pi \over 2}} {\cos xdx} + \int\limits_{{\pi \over 2}}^\pi { - \cos xdx} } \right] + \int\limits_0^{{\pi \over 2}} {\cos xdx} + \int\limits_{{\pi \over 2}}^v { - \cos xdx} $$ $$I = n\left[ {\left( {\sin x} \right)_0^{{\pi \over 2}} - \left( {\sin x} \right)_{{\pi \over 2}}^\pi } \right] + \left[ {\sin x} \right]_0^{{\pi \over 2}} - \left[ {\sin x} \right]_{{\pi \over 2}}^v$$ $$I = n\left[ {1 - 0 - 0 + 1} \right] + 1 - 0 - \sin v + 1$$ $$I = 2n + 2 - \sin v$$