The area enclosed between the curves

$${y^2} = x$$ and $$y = \left| x \right|$$

From the figure, area lies between

$${y^2} = x$$ and $$y = x$$



$$\therefore$$ Required area $$ = \int_0^1 {\left( {{y_2} - {y_1}} \right)} dx$$

$$ = \int_0^1 {\left( {\sqrt x - x} \right)dx = \left[ {{{{x^{3/2}}} \over {3/2}} - {{{x^2}} \over 2}} \right]} _0^1$$

$$\therefore$$ Required area $$ = {2 \over 3}\left[ {{x^{3/2}}} \right]_0^1 - {1 \over 2}\left[ {{x^2}} \right]_0^1$$

$$ = {2 \over 3} - {1 \over 2} = {1 \over 6}$$

$$\int_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }}} = {\pi \over 2}$$

$$\therefore$$ $$\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}$$

$$\left[ {} \right.$$ As $$\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x$$ $$\left. {} \right]$$

$$ \Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = {\pi \over 2}$$

$$ \Rightarrow {\sec ^{ - 1}}x - {\pi \over 4} = {\pi \over 2}$$

$$ \Rightarrow {\sec ^{ - 1}}x = {\pi \over 2} + {\pi \over 4}$$

$$ \Rightarrow {\sec ^{ - 1}}x = {{3\pi } \over 4}$$

$$ \Rightarrow x = \sec {{3\pi } \over 4}$$

$$ \Rightarrow x = - \sqrt 2 $$

Given $$f\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$

where $$f\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt$$

$$\therefore$$ $$F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)$$

$$ \Rightarrow F\left( e \right)$$

$$ = \int_1^e {{{\log \,t} \over {1 + t}}dt + \int_1^{1/e} {{{\log t} \over {1 + t}}} } dt\,\,\,....\left( A \right)$$

Now for solving, $$I = \int_1^{1/e} {{{\log t} \over {1 + t}}dt} $$

$$\therefore$$ Put $${1 \over t} = z \Rightarrow - {1 \over {{t^2}}}dt = dz$$

$$ \Rightarrow dt = - {{dz} \over {{z^2}}}$$

and limit for $$t = 1 \Rightarrow z = 1$$ and for

$$t = 1/e \Rightarrow z = e$$

$$\therefore$$ $$I = \int_1^e {{{\log \left( {{1 \over z}} \right)} \over {1 + {1 \over z}}}} \left( { - {{dz} \over {{z^2}}}} \right)$$

$$ = \int_1^e {{{\left( {\log 1 - \log z} \right).z} \over {z + 1}}\left( { - {{dz} \over {{z^2}}}} \right)} $$

$$ = \int_1^e { - {{\log z} \over {\left( {z + 1} \right)}}\left( { - {{dz} \over z}} \right)} $$

[ as $$\log 1 = 0$$ ]

$$ = \int_1^e {{{\log z} \over {z\left( {z + 1} \right)}}} dz$$

$$\therefore$$ $$I = \int_1^e {{{\log \,t} \over {t\left( {t + 1} \right)}}dt} $$

$$\left[ {} \right.$$ By property $$\int_a^b {f\left( t \right)dt} $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \int_a^b {f\left( x \right)dx} $$ $$\left. {} \right]$$

Equation $$(A)$$ becomes

$$F\left( e \right) = \int_1^e {{{\log t} \over {1 + t}}dt + \int_1^e {{{\log t} \over {t\left( {1 + t} \right)}}} } dt$$

$$ = \int_1^e {{{t.\log t + \log t} \over {t\left( {1 + t} \right)}}} dt$$

$$ = \int_1^e {{{\left( {\log t} \right)\left( {t + 1} \right)} \over {t\left( {1 + t} \right)}}} $$

$$ \Rightarrow F\left( e \right) = \int_1^e {{{\log t} \over t}dt} $$

Let $$\log t = x$$

$$\therefore$$ $${1 \over t}dt = dx$$

$$\left[ {} \right.$$ for limit $$t = 1,x = 0$$

and $$t = e,x = \log \,e = 1$$ $$\left. {} \right]$$

$$\therefore$$ $$F\left( e \right) = \int_0^1 {x\,dx} $$

$$ \Rightarrow F\left( e \right) = \left[ {{{{x^2}} \over 2}} \right]_0^1$$

$$ \Rightarrow F\left( e \right) = {1 \over 2}$$

Let $$a = k + h$$ where $$k$$ is an integer such that

$$\left[ a \right] = k$$ and $$0 \le h < 1$$

$$\therefore$$ $$\int\limits_1^a {\left[ x \right]f'\left( x \right)dx = \int\limits_1^2 {1f'\left( x \right)} } \,dx$$

$$\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_2^3 {2f'\left( x \right)dx + } ....$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{k - 1}^k {\left( {k - 1} \right)dx + \int\limits_k^{k + h} {kf'\left( x \right)} } dx$$

$$\left\{ {f\left( 2 \right) - f\left( 1 \right)} \right\} + 2\left\{ {f\left( 3 \right) - f\left( 2 \right)} \right\}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ + 3\left\{ {f\left( 4 \right) - f\left( 3 \right)} \right\} + \,........\,$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ + \left( {k - 1} \right)\left\{ {f\left( k \right) - f\left( {k - 1} \right)} \right\}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ + k\left\{ {f\left( {k + h} \right) - f\left( k \right)} \right\}$$

$$ = - f\left( 1 \right) - f\left( 2 \right) - f\left( 3 \right).......$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ - f\left( k \right) + kf\left( {k + h} \right)$$

$$ = \left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right)} \right.$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\left. { + f\left( 3 \right) + ........f\left( {\left[ a \right]} \right)} \right\}$$