put \( z=\mathrm{Re}^{i \theta} \), then \( |z|=R \) Now, \( \left|\int_{\Gamma} f(z) d z\right| \leq \frac{\pi R}{\left(R^{2}-1\right)^{2}} \rightarrow 0 \) as \( R \rightarrow \infty \) Thus, the equation (1) becomes \( \lim _{R \rightarrow \infty} \int_{-R}^{R} f(x) d x+\lim _{R \rightarrow \infty} \int_{\Gamma} f(z) d z=\frac{\pi}{2} \) \( \Rightarrow \quad \int_{-\infty}^{\infty} f(x) d x+0=\frac{\pi}{2} \Rightarrow \int_{-\infty}^{\infty} f(x) d x=\frac{\pi}{2} \)

5) A particle is moving in a straight line with velocity \( v(t)=15 t^{2}+2 \). Find; a) Its acceleration after 3 seconds b) Its displacement after 2 secnds

4) The acceleration of a particle moving in a straight line at a time \( t \) seconds, is given by \( a(t)=t+1 \). Given that \( s=0 \) and \( v=8 \) when \( t=0 \), find formula for: (a) The velocity \( v \) (b) The displacement

25. Given \( f(x)=\left\{\begin{array}{c}x+4 \text { if } x<3 \\ 2 x^{2}-x-1 \text { if } x \geq 3\end{array}\right. \) and \( g(x)=\left\{\begin{array}{c}x \text { if } x<-2 \\ 2 x+5 \text { if }-2 \leq x<5 . \\ x^{2}-1 \text { if } x \geq 5\end{array}\right. \) What must be the value of \( a \) so that \( (f \circ g)(a)=\left(8 a^{2}+40 a+50\right)-2 a-6 \) ? A. \( a \) is a value greater than 5. B. \( a \) is any value greater or equal to 3. C. \( a \) is a value between 3 and 5. D. \( a \) is a value smaller than 3. E. \( a \) is a value smaller than -2.

seconds, is given by \( a(t)=t+1 \). Given that \( s=0 \) and \( v=8 \) when \( t=0 \), find formula for: (a) The velocity \( v \) (b) The displacement

\( F ( x ) = \int _ { 0 } ^ { \sin x } \frac { d t } { 1 + t ^ { 3 } } \)