We have,
\[\left| x \right| + \left| {x - 1} \right| = \left\{ \begin{array}{l} - x - \left( {x - 1} \right) = - 2x + 1,\,{\rm{if}}\,x \le 0\\ \,\,\,\,\,x - \left( {x - 1} \right) = 1,\,{\rm{if}}\,0 \le x \le 1\\ \,\,\,\,\,x + x - 1 = 2x - 1,\,{\rm{if}}\,x \ge 1 \end{array} \right.\]
$$\eqalign{ & \therefore \,\int_{ - 1}^3 {\left( {\left| x \right| + \left| {x - 1} \right|} \right)dx} \cr & = \int_{ - 1}^0 {\left( { - 2x + 1} \right)dx} + \int_0^1 {1\,dx} + \int_1^3 {\left( {2x - 1} \right)dx} \cr & = \left[ { - {x^2} + x} \right]_{ - 1}^0 + \left[ x \right]_0^1 + \left[ {{x^2} - x} \right]_1^3 \cr & = 9 \cr} $$

$$\eqalign{ & \int_0^\pi {{{\sec }^2}x\,dx} = \mathop {\lim }\limits_{ \in \to 0} \left( {\int_0^{\frac{\pi }{2} - \in } {{{\sec }^2}x\,dx + } \int_{\frac{\pi }{2} + \in }^\pi {{{\sec }^2}x\,dx} } \right) \cr & = \mathop {\lim }\limits_{ \in \to 0} \left\{ {\tan \left( {\frac{\pi }{2} - \in } \right) - \tan \left( {\frac{\pi }{2} + \in } \right)} \right\} \cr & = \mathop {\lim }\limits_{ \in \to 0} \left( {\cot \, \in + \cot \in } \right) \cr & = \infty \cr} $$

We have
$$\eqalign{ & {R_1} = \int_{ - 1}^2 {x\,f\left( x \right)} dx = \int_{ - 1}^2 {\left( {1 - x} \right)f\left( {1 - x} \right)dx} \cr & \left[ {{\text{Using }}\int_a^b {f\left( x \right)dx = \int_a^b {f\left( {a + b - x} \right)dx} } } \right] \cr & \Rightarrow {R_1} = \int_{ - 1}^2 {\left( {1 - x} \right)f\left( x \right)dx} \cr & \left[ {{\text{As}}\,{\text{ }}f\left( x \right) = f\left( {1 - x} \right){\text{ on }}\left[ { - 1,\,2} \right]} \right] \cr & \therefore {R_1} + {R_2} = \int_{ - 1}^2 {x\,f\left( x \right)dx + } \int_{ - 1}^2 {\left( {1 - x} \right)f\left( x \right)dx} \cr & \Rightarrow 2{R_1} = \int_{ - 1}^2 {f\left( x \right)dx = {R_2}} \cr} $$

As $$f\left( x \right)$$  satisfies the conditions of Rolle’s theorem in [1, 2], $$f\left( x \right)$$  is continuous in the interval and $$f\left( 1 \right) = f\left( 2 \right).$$
$$\therefore \int_1^2 {f'\left( x \right)} dx = \left[ {f\left( x \right)} \right]_1^2 = f\left( 2 \right) - f\left( 1 \right) = 0$$

$$\eqalign{ & {\text{Put }}z = \frac{1}{t} \cr & {\text{Then }}I = \int_x^{\frac{1}{x}} {f\left( {\frac{1}{t}} \right).\frac{{ - 1}}{{{t^2}}}\,} dt \cr & = \int_x^{\frac{1}{x}} {\left\{ { - {t^2}.f\left( t \right)} \right\}.\frac{{ - 1}}{{{t^2}}}dt} \,\,\left( {{\text{from the question}}} \right) \cr & = \int_x^{\frac{1}{x}} {f\left( t \right)dt} \cr & = - \int_{\frac{1}{x}}^x {f\left( t \right)dt} \cr & = - I \cr & \therefore \,\,2I = 0 \cr} $$

$$\eqalign{ & F'\left( x \right) = f\left( x \right) \cr & {\text{Also, }}F\left( t \right) = t\left( {1 + \sqrt t } \right) \cr & \Rightarrow F'\left( t \right) = 1 + \frac{3}{2}{t^{\frac{1}{2}}}\,; \cr & F'\left( 4 \right) = 1 + 3 = 4 \Rightarrow f\left( 4 \right) = 4 \cr} $$

$$\eqalign{ & {\text{Here}}\,\,f'\left( x \right) = \phi \left( x \right) \cr & {\text{So}},\,\,\int_a^\lambda {\phi \left( x \right)dx = \int_a^\lambda {f'\left( x \right)dx} } = \left[ {f\left( x \right)} \right]_a^\lambda = f\left( \lambda \right) - f\left( a \right) \cr & {\text{But}}\,\,b = f\left( a \right),\,\,\mu = f\left( \lambda \right).\,\,{\text{So}},\,\,\int_a^\lambda {\phi \left( x \right)dx = \mu - } b \cr} $$

$$I = \int\limits_0^{10\pi } {\left| {\sin \,x\,} \right|dx} = 10\int\limits_0^\pi {\left| {\sin \,x\,} \right|dx} = 10\int\limits_0^\pi {\sin \,x\,dx} $$
[$$\because \left| {\sin \,x} \right|$$   is periodic with period $$\pi $$ and $$\sin \,x > 0$$   if $$0 < x < \pi $$   ]
$$I = 20\int\limits_0^{\frac{\pi }{2}} {\sin \,x\,dx = 20\left[ { - \cos \,x} \right]_0^{\frac{\pi }{2}} = 20} $$

$$\eqalign{ & I = \int_{\frac{\pi }{3}}^{\frac{\pi }{2}} {\sin \,x.d\left( {{x^3}} \right)} = 3\int_{\frac{\pi }{3}}^{\frac{\pi }{2}} {{x^2}\sin \,x\,dx} \cr & \int {{x^2}\sin \,x\,dx} = {x^2}\left( { - \cos \,x} \right) + \int {2x\cos \,x\,dx} \cr & = - {x^2}\cos \,x + 2\left\{ {x\sin \,x - \int {\sin \,x\,dx} } \right\} \cr & = - {x^2}\cos \,x + 2x\sin \,x + 2\cos \,x \cr & \therefore I = 3\left[ { - {x^2}\cos \,x + 2x\sin \,x + 2\cos \,x} \right]_{\frac{\pi }{3}}^{\frac{\pi }{2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 3\pi - 3\left( { - \frac{{{\pi ^2}}}{9}.\frac{1}{2} + 2.\frac{\pi }{3}.\frac{{\sqrt 3 }}{2} + 2.\frac{1}{2}} \right) \cr & = 3\pi + \frac{{{\pi ^2}}}{6} - \sqrt 3 \pi - 3 \cr & = \frac{{{\pi ^2}}}{6} + \left( {3 - \sqrt 3 } \right)\pi - 3 \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 1} \int\limits_0^{f\left( x \right)} {\frac{{2t}}{{x - 1}}dt} \cr & = \mathop {\lim }\limits_{x \to 1} \frac{1}{{x - 1}}\left[ {{t^2}} \right]_0^{f\left( x \right)} \cr & = \mathop {\lim }\limits_{x \to 1} \frac{{{f^2}\left( x \right) - {f^2}\left( 0 \right)}}{{x - 1}} \cr & = \mathop {\lim }\limits_{x \to 1} \frac{{2f\left( x \right)f'\left( x \right)}}{1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{using L'Hospital Rule}}} \right) \cr & = 2f\left( 1 \right)f'\left( 1 \right) \cr & = 8f'\left( 1 \right) \cr} $$