\displaystyle \text{Evaluate the following integrals:}

\displaystyle \textbf{Question 1: }~
\displaystyle \text{(i): }\int_{1}^{4} f(x)\,dx,\ \text{where }f(x)=\begin{cases}4x+3,&1\le x\le 2\\[4pt]3x+5,&2\le x\le 4\end{cases}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \int_{1}^{4} f(x)\,dx,\ \text{where } f(x)=\begin{cases}4x+3, & 1\le x\le 2\\ 3x+5, & 2\le x\le 4\end{cases}
\displaystyle I=\int_{1}^{4} f(x)\,dx
\displaystyle \Rightarrow I=\int_{1}^{2} f(x)\,dx+\int_{2}^{4} f(x)\,dx
\displaystyle \Rightarrow I=\int_{1}^{2} (4x+3)\,dx+\int_{2}^{4} (3x+5)\,dx
\displaystyle \Rightarrow I=\left[2x^{2}+3x\right]_{1}^{2}+\left[\frac{3x^{2}}{2}+5x\right]_{2}^{4}
\displaystyle \Rightarrow I=\left(8+6-2-3\right)+\left(24+20-6-10\right)
\displaystyle \Rightarrow I=37

\displaystyle \text{(ii): }\int_{0}^{9} f(x)\,dx,\ \text{where }f(x)=\begin{cases}\sin x,&0\le x\le \pi/2\\[4pt]1,&\pi/2\le x\le 3\\[4pt]e^{x-3},&3\le x\le 9\end{cases}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \int_{0}^{9} f(x)\,dx,\ \text{where } f(x)=\begin{cases}\sin x, & 0\le x\le \frac{\pi}{2}\\ 1, & \frac{\pi}{2}\le x\le 3\\ e^{x-3}, & 3\le x\le 9\end{cases}
\displaystyle I=\int_{0}^{9} f(x)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{2}} f(x)\,dx+\int_{\frac{\pi}{2}}^{3} f(x)\,dx+\int_{3}^{9} f(x)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{2}} \sin x\,dx+\int_{\frac{\pi}{2}}^{3} 1\,dx+\int_{3}^{9} e^{x-3}\,dx
\displaystyle \Rightarrow I=\left[-\cos x\right]_{0}^{\frac{\pi}{2}}+\left[x\right]_{\frac{\pi}{2}}^{3}+\left[e^{x-3}\right]_{3}^{9}
\displaystyle \Rightarrow I=\left(0-(-1)\right)+\left(3-\frac{\pi}{2}\right)+\left(e^{6}-e^{0}\right)
\displaystyle \Rightarrow I=3-\frac{\pi}{2}+e^{6}

\displaystyle \text{(iii): }\int_{1}^{4} f(x)\,dx,\ \text{where }f(x)=\begin{cases}7x+3,&1\le x\le 3\\[4pt]8x,&3\le x\le 4\end{cases}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \int_{1}^{4} f(x)\,dx,\ \text{where } f(x)=\begin{cases}7x+3, & 1\le x\le 3\\ 8x, & 3\le x\le 4\end{cases}
\displaystyle I=\int_{1}^{4} f(x)\,dx
\displaystyle \Rightarrow I=\int_{1}^{3} f(x)\,dx+\int_{3}^{4} f(x)\,dx
\displaystyle \Rightarrow I=\int_{1}^{3} (7x+3)\,dx+\int_{3}^{4} 8x\,dx
\displaystyle \Rightarrow I=\left[\frac{7x^{2}}{2}+3x\right]_{1}^{3}+\left[4x^{2}\right]_{3}^{4}
\displaystyle \Rightarrow I=\left(\frac{63}{2}+9-\frac{7}{2}-3\right)+\left(64-36\right)
\displaystyle \Rightarrow I=\frac{56}{2}+6+28
\displaystyle \Rightarrow I=62

\displaystyle \textbf{Question 2: }~\int_{-4}^{4}\lvert x+2\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_{-4}^{4} |x+2|\,dx
\displaystyle \text{We know that } |x+2|=\begin{cases}-(x+2), & -4\le x\le -2\\ x+2, & -2\le x\le 4\end{cases}
\displaystyle \therefore I=\int_{-4}^{4} |x+2|\,dx
\displaystyle \Rightarrow I=\int_{-4}^{-2} -(x+2)\,dx+\int_{-2}^{4} (x+2)\,dx
\displaystyle \Rightarrow I=\left[-\left(\frac{x^{2}}{2}+2x\right)\right]_{-4}^{-2}+\left[\frac{x^{2}}{2}+2x\right]_{-2}^{4}
\displaystyle \Rightarrow I=\left(\left[-\left(\frac{x^{2}}{2}+2x\right)\right]_{x=-2}-\left[-\left(\frac{x^{2}}{2}+2x\right)\right]_{x=-4}\right)+\left(\left[\frac{x^{2}}{2}+2x\right]_{x=4}-\left[\frac{x^{2}}{2}+2x\right]_{x=-2}\right)
\displaystyle \Rightarrow I=\left(-\left(\frac{4}{2}-4\right)+\left(\frac{16}{2}-8\right)\right)+\left(\left(\frac{16}{2}+8\right)-\left(\frac{4}{2}-4\right)\right)
\displaystyle \Rightarrow I=\left(-\left(2-4\right)+\left(8-8\right)\right)+\left(\left(8+8\right)-\left(2-4\right)\right)
\displaystyle \Rightarrow I=\left(-(-2)+0\right)+\left(16-(-2)\right)
\displaystyle \Rightarrow I=2+18
\displaystyle \Rightarrow I=20

\displaystyle \textbf{Question 3: }~\int_{-3}^{3}\lvert x+1\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-3}^{3} |x+1|\,dx
\displaystyle \text{We know that } |x+1|=\begin{cases}-(x+1), & -3\le x\le -1\\ x+1, & -1\le x\le 3\end{cases}
\displaystyle \therefore I=\int_{-3}^{3} |x+1|\,dx
\displaystyle \Rightarrow I=\int_{-3}^{-1} -(x+1)\,dx+\int_{-1}^{3} (x+1)\,dx
\displaystyle \Rightarrow I=\left[-\left(\frac{(x+1)^{2}}{2}\right)\right]_{-3}^{-1}+\left[\frac{(x+1)^{2}}{2}\right]_{-1}^{3}
\displaystyle \Rightarrow I=\left(\left[-\frac{(x+1)^{2}}{2}\right]_{x=-1}-\left[-\frac{(x+1)^{2}}{2}\right]_{x=-3}\right)+\left(\left[\frac{(x+1)^{2}}{2}\right]_{x=3}-\left[\frac{(x+1)^{2}}{2}\right]_{x=-1}\right)
\displaystyle \Rightarrow I=\left(-\frac{0}{2}-\left(-\frac{4}{2}\right)\right)+\left(\frac{16}{2}-\frac{0}{2}\right)
\displaystyle \Rightarrow I=2+8
\displaystyle \Rightarrow I=10

\displaystyle \textbf{Question 4: }~\int_{-1}^{1}\lvert 2x+1\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-1}^{1} |2x+1|\,dx
\displaystyle \text{We know that } |2x+1|=\begin{cases}-(2x+1), & -1\le x\le -\frac{1}{2}\\ 2x+1, & -\frac{1}{2}\le x\le 1\end{cases}
\displaystyle \therefore I=\int_{-1}^{1} |2x+1|\,dx
\displaystyle \Rightarrow I=\int_{-1}^{-\frac{1}{2}} -(2x+1)\,dx+\int_{-\frac{1}{2}}^{1} (2x+1)\,dx
\displaystyle \Rightarrow I=\left[-\left(x^{2}+x\right)\right]_{-1}^{-\frac{1}{2}}+\left[x^{2}+x\right]_{-\frac{1}{2}}^{1}
\displaystyle \Rightarrow I=\left(\left[-\left(x^{2}+x\right)\right]_{x=-\frac{1}{2}}-\left[-\left(x^{2}+x\right)\right]_{x=-1}\right)+\left(\left[x^{2}+x\right]_{x=1}-\left[x^{2}+x\right]_{x=-\frac{1}{2}}\right)
\displaystyle \Rightarrow I=\left(-\left(\frac{1}{4}-\frac{1}{2}\right)+\left(1-1\right)\right)+\left(\left(1+1\right)-\left(\frac{1}{4}-\frac{1}{2}\right)\right)
\displaystyle \Rightarrow I=\left(-\left(-\frac{1}{4}\right)+0\right)+\left(2-\left(-\frac{1}{4}\right)\right)
\displaystyle \Rightarrow I=\frac{1}{4}+\frac{9}{4}
\displaystyle \Rightarrow I=\frac{5}{2}

\displaystyle \textbf{Question 5: }~\int_{-2}^{2}\lvert 2x+3\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-2}^{2} |2x+3|\,dx
\displaystyle \text{We know that } |2x+3|=\begin{cases}-(2x+3), & -2\le x\le -\frac{3}{2}\\ 2x+3, & -\frac{3}{2}\le x\le 2\end{cases}
\displaystyle \therefore I=\int_{-2}^{2} |2x+3|\,dx
\displaystyle \Rightarrow I=\int_{-2}^{-\frac{3}{2}} -(2x+3)\,dx+\int_{-\frac{3}{2}}^{2} (2x+3)\,dx
\displaystyle \Rightarrow I=\left[-\left(x^{2}+3x\right)\right]_{-2}^{-\frac{3}{2}}+\left[x^{2}+3x\right]_{-\frac{3}{2}}^{2}
\displaystyle \Rightarrow I=\left(\left[-\left(x^{2}+3x\right)\right]_{x=-\frac{3}{2}}-\left[-\left(x^{2}+3x\right)\right]_{x=-2}\right)+\left(\left[x^{2}+3x\right]_{x=2}-\left[x^{2}+3x\right]_{x=-\frac{3}{2}}\right)
\displaystyle \Rightarrow I=\left(-\left(\frac{9}{4}-\frac{9}{2}\right)+\left(4-6\right)\right)+\left(\left(4+6\right)-\left(\frac{9}{4}-\frac{9}{2}\right)\right)
\displaystyle \Rightarrow I=\left(-\left(-\frac{9}{4}\right)-2\right)+\left(10-\left(-\frac{9}{4}\right)\right)
\displaystyle \Rightarrow I=\frac{9}{4}-2+\frac{49}{4}
\displaystyle \Rightarrow I=\frac{25}{2}

\displaystyle \textbf{Question 6: }~\int_{0}^{2}\lvert x^2-3x+2\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{2} |x^{2}-3x+2|\,dx
\displaystyle \text{We know that } |x^{2}-3x+2|=\begin{cases}x^{2}-3x+2, & 0\le x\le 1\\ -(x^{2}-3x+2), & 1\le x\le 2\end{cases}
\displaystyle \therefore I=\int_{0}^{2} |x^{2}-3x+2|\,dx
\displaystyle \Rightarrow I=\int_{0}^{1} (x^{2}-3x+2)\,dx+\int_{1}^{2} -(x^{2}-3x+2)\,dx
\displaystyle \Rightarrow I=\left[\frac{x^{3}}{3}-\frac{3x^{2}}{2}+2x\right]_{0}^{1}-\left[\frac{x^{3}}{3}-\frac{3x^{2}}{2}+2x\right]_{1}^{2}
\displaystyle \Rightarrow I=\left(\frac{1}{3}-\frac{3}{2}+2\right)-\left(\frac{8}{3}-6+4-\frac{1}{3}+\frac{3}{2}-2\right)
\displaystyle \Rightarrow I=\frac{1}{3}-\frac{3}{2}+2-\frac{8}{3}+6-4+\frac{1}{3}-\frac{3}{2}+2
\displaystyle \Rightarrow I=1

\displaystyle \textbf{Question 7: }~\int_{0}^{3}\lvert 3x-1\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{3} |3x-1|\,dx
\displaystyle \text{We know that } |3x-1|=\begin{cases}-(3x-1), & 0\le x\le \frac{1}{3}\\ 3x-1, & \frac{1}{3}\le x\le 3\end{cases}
\displaystyle \therefore I=\int_{0}^{3} |3x-1|\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{1}{3}} -(3x-1)\,dx+\int_{\frac{1}{3}}^{3} (3x-1)\,dx
\displaystyle \Rightarrow I=\left[-\left(\frac{3x^{2}}{2}-x\right)\right]_{0}^{\frac{1}{3}}+\left[\frac{3x^{2}}{2}-x\right]_{\frac{1}{3}}^{3}
\displaystyle \Rightarrow I=\left(\left[-\left(\frac{3x^{2}}{2}-x\right)\right]_{x=\frac{1}{3}}-\left[-\left(\frac{3x^{2}}{2}-x\right)\right]_{x=0}\right)+\left(\left[\frac{3x^{2}}{2}-x\right]_{x=3}-\left[\frac{3x^{2}}{2}-x\right]_{x=\frac{1}{3}}\right)
\displaystyle \Rightarrow I=\left(-\left(\frac{1}{6}-\frac{1}{3}\right)-0\right)+\left(\left(\frac{27}{2}-3\right)-\left(\frac{1}{6}-\frac{1}{3}\right)\right)
\displaystyle \Rightarrow I=\left(-\left(-\frac{1}{6}\right)\right)+\left(\frac{27}{2}-3+\frac{1}{6}\right)
\displaystyle \Rightarrow I=\frac{1}{6}+\frac{63}{6}
\displaystyle \Rightarrow I=\frac{65}{6}

\displaystyle \textbf{Question 8: }~\int_{-6}^{6}\lvert x+2\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-6}^{6} |x+2|\,dx
\displaystyle \text{We know that } |x+2|=\begin{cases}-(x+2), & -6\le x\le -2\\ x+2, & -2\le x\le 6\end{cases}
\displaystyle \therefore I=\int_{-6}^{6} |x+2|\,dx
\displaystyle \Rightarrow I=\int_{-6}^{-2} -(x+2)\,dx+\int_{-2}^{6} (x+2)\,dx
\displaystyle \Rightarrow I=\left[-\left(\frac{x^{2}}{2}+2x\right)\right]_{-6}^{-2}+\left[\frac{x^{2}}{2}+2x\right]_{-2}^{6}
\displaystyle \Rightarrow I=\left(\left[-\left(\frac{x^{2}}{2}+2x\right)\right]_{x=-2}-\left[-\left(\frac{x^{2}}{2}+2x\right)\right]_{x=-6}\right)+\left(\left[\frac{x^{2}}{2}+2x\right]_{x=6}-\left[\frac{x^{2}}{2}+2x\right]_{x=-2}\right)
\displaystyle \Rightarrow I=\left(-\left(2-4\right)+\left(18-12\right)\right)+\left(\left(18+12\right)-\left(2-4\right)\right)
\displaystyle \Rightarrow I=\left(-(-2)+6\right)+\left(30-(-2)\right)
\displaystyle \Rightarrow I=8+32
\displaystyle \Rightarrow I=40

\displaystyle \textbf{Question 9: }~\int_{-2}^{2}\lvert x+1\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-2}^{2} |x+1|\,dx
\displaystyle \text{We know that } |x+1|=\begin{cases}-(x+1), & -2\le x\le -1\\ x+1, & -1\le x\le 2\end{cases}
\displaystyle \therefore I=\int_{-2}^{2} |x+1|\,dx
\displaystyle \Rightarrow I=\int_{-2}^{-1} -(x+1)\,dx+\int_{-1}^{2} (x+1)\,dx
\displaystyle \Rightarrow I=\left[-\left(\frac{x^{2}}{2}+x\right)\right]_{-2}^{-1}+\left[\frac{x^{2}}{2}+x\right]_{-1}^{2}
\displaystyle \Rightarrow I=\left(\left[-\left(\frac{x^{2}}{2}+x\right)\right]_{x=-1}-\left[-\left(\frac{x^{2}}{2}+x\right)\right]_{x=-2}\right)+\left(\left[\frac{x^{2}}{2}+x\right]_{x=2}-\left[\frac{x^{2}}{2}+x\right]_{x=-1}\right)
\displaystyle \Rightarrow I=\left(-\left(\frac{1}{2}-1\right)+\left(2-2\right)\right)+\left(\left(2+2\right)-\left(\frac{1}{2}-1\right)\right)
\displaystyle \Rightarrow I=\left(-\left(-\frac{1}{2}\right)+0\right)+\left(4-\left(-\frac{1}{2}\right)\right)
\displaystyle \Rightarrow I=\frac{1}{2}+\frac{9}{2}
\displaystyle \Rightarrow I=5

\displaystyle \textbf{Question 10: }~\int_{1}^{2}\lvert x-3\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{1}^{2} |x-3|\,dx
\displaystyle \text{We know that } |x-3|=\begin{cases}3-x, & 1\le x\le 2\end{cases}
\displaystyle \therefore I=\int_{1}^{2} |x-3|\,dx
\displaystyle \Rightarrow I=\int_{1}^{2} (3-x)\,dx
\displaystyle \Rightarrow I=\left[3x-\frac{x^{2}}{2}\right]_{1}^{2}
\displaystyle \Rightarrow I=\left(6-2\right)-\left(3-\frac{1}{2}\right)
\displaystyle \Rightarrow I=4-\frac{5}{2}
\displaystyle \Rightarrow I=\frac{3}{2}

\displaystyle \textbf{Question 11: }~\int_{0}^{\pi/2}\lvert \cos 2x\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{\frac{\pi}{2}} |\cos 2x|\,dx
\displaystyle \text{We know that } |\cos 2x|=\begin{cases}\cos 2x, & 0\le x\le \frac{\pi}{4}\\ -\cos 2x, & \frac{\pi}{4}\le x\le \frac{\pi}{2}\end{cases}
\displaystyle \therefore I=\int_{0}^{\frac{\pi}{2}} |\cos 2x|\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{4}} \cos 2x\,dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (-\cos 2x)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{4}} \cos 2x\,dx-\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos 2x\,dx
\displaystyle \Rightarrow I=\left[\frac{\sin 2x}{2}\right]_{0}^{\frac{\pi}{4}}-\left[\frac{\sin 2x}{2}\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}
\displaystyle \Rightarrow I=\left(\frac{\sin\left(\frac{\pi}{2}\right)}{2}-\frac{\sin 0}{2}\right)-\left(\frac{\sin(\pi)}{2}-\frac{\sin\left(\frac{\pi}{2}\right)}{2}\right)
\displaystyle \Rightarrow I=\left(\frac{1}{2}-0\right)-\left(0-\frac{1}{2}\right)
\displaystyle \Rightarrow I=1

\displaystyle \textbf{Question 12: }~\int_{0}^{2\pi}\lvert \sin x\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{2\pi} |\sin x|\,dx
\displaystyle \text{We know that } |\sin x|=\begin{cases}\sin x, & 0\le x\le \pi\\ -\sin x, & \pi\le x\le 2\pi\end{cases}
\displaystyle \therefore I=\int_{0}^{2\pi} |\sin x|\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi} \sin x\,dx+\int_{\pi}^{2\pi} (-\sin x)\,dx
\displaystyle \Rightarrow I=\left[-\cos x\right]_{0}^{\pi}-\left[-\cos x\right]_{\pi}^{2\pi}
\displaystyle \Rightarrow I=\left(-\cos\pi+\cos 0\right)-\left(-\cos 2\pi+\cos\pi\right)
\displaystyle \Rightarrow I=\left(1+1\right)-\left(-1-1\right)
\displaystyle \Rightarrow I=4

\displaystyle \textbf{Question 13: }~\int_{-\pi/4}^{\pi/4}\lvert \sin x\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin x|\,dx
\displaystyle \text{We know that } |\sin x|=\begin{cases}-\sin x, & -\frac{\pi}{4}\le x\le 0\\ \sin x, & 0\le x\le \frac{\pi}{4}\end{cases}
\displaystyle \therefore I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin x|\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{4}}^{0} (-\sin x)\,dx+\int_{0}^{\frac{\pi}{4}} \sin x\,dx
\displaystyle \Rightarrow I=\left[\cos x\right]_{-\frac{\pi}{4}}^{0}+\left[-\cos x\right]_{0}^{\frac{\pi}{4}}
\displaystyle \Rightarrow I=\left(\cos 0-\cos\left(-\frac{\pi}{4}\right)\right)+\left(-\cos\left(\frac{\pi}{4}\right)+\cos 0\right)
\displaystyle \Rightarrow I=\left(1-\frac{1}{\sqrt{2}}\right)+\left(-\frac{1}{\sqrt{2}}+1\right)
\displaystyle \Rightarrow I=2-\frac{2}{\sqrt{2}}
\displaystyle \Rightarrow I=2-\sqrt{2}

\displaystyle \textbf{Question 14: }~\int_{2}^{8}\lvert x-5\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{2}^{8} |x-5|\,dx
\displaystyle \text{We know that } |x-5|=\begin{cases}-(x-5), & 2\le x\le 5\\ x-5, & 5\le x\le 8\end{cases}
\displaystyle \therefore I=\int_{2}^{8} |x-5|\,dx
\displaystyle \Rightarrow I=\int_{2}^{5} -(x-5)\,dx+\int_{5}^{8} (x-5)\,dx
\displaystyle \Rightarrow I=\left[-\left(\frac{x^{2}}{2}-5x\right)\right]_{2}^{5}+\left[\frac{x^{2}}{2}-5x\right]_{5}^{8}
\displaystyle \Rightarrow I=\left(\left[-\left(\frac{x^{2}}{2}-5x\right)\right]_{x=5}-\left[-\left(\frac{x^{2}}{2}-5x\right)\right]_{x=2}\right)+\left(\left[\frac{x^{2}}{2}-5x\right]_{x=8}-\left[\frac{x^{2}}{2}-5x\right]_{x=5}\right)
\displaystyle \Rightarrow I=\left(-\left(\frac{25}{2}-25\right)+\left(2-10\right)\right)+\left(\left(32-40\right)-\left(\frac{25}{2}-25\right)\right)
\displaystyle \Rightarrow I=\left(-\left(-\frac{25}{2}\right)-8\right)+\left(-8-\left(-\frac{25}{2}\right)\right)
\displaystyle \Rightarrow I=\left(\frac{25}{2}-8\right)+\left(-8+\frac{25}{2}\right)
\displaystyle \Rightarrow I=\frac{9}{2}+\frac{9}{2}
\displaystyle \Rightarrow I=9

\displaystyle \textbf{Question 15: }~\int_{-\pi/2}^{\pi/2}\big(\sin\lvert x\rvert+\cos\lvert x\rvert\big)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\sin|x|+\cos|x|\right)\,dx
\displaystyle \text{Since } f(-x)=\sin|-x|+\cos|-x|=\sin|x|+\cos|x|=f(x)
\displaystyle \text{So, } f(x)\text{ is an even function.}
\displaystyle \therefore I=2\int_{0}^{\frac{\pi}{2}} \left(\sin x+\cos x\right)\,dx
\displaystyle \Rightarrow I=2\left[-\cos x+\sin x\right]_{0}^{\frac{\pi}{2}}
\displaystyle \Rightarrow I=2\left((0+1)-(-1+0)\right)
\displaystyle \Rightarrow I=2(2)
\displaystyle \Rightarrow I=4

\displaystyle \textbf{Question 16: }~\int_{0}^{4}\lvert x-1\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{4} |x-1|\,dx
\displaystyle \text{We know that } |x-1|=\begin{cases}-(x-1), & 0\le x\le 1\\ x-1, & 1\le x\le 4\end{cases}
\displaystyle \therefore I=\int_{0}^{4} |x-1|\,dx
\displaystyle \Rightarrow I=\int_{0}^{1} -(x-1)\,dx+\int_{1}^{4} (x-1)\,dx
\displaystyle \Rightarrow I=\left[-\left(\frac{x^{2}}{2}-x\right)\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{4}
\displaystyle \Rightarrow I=\left(\left[-\left(\frac{x^{2}}{2}-x\right)\right]_{x=1}-\left[-\left(\frac{x^{2}}{2}-x\right)\right]_{x=0}\right)+\left(\left[\frac{x^{2}}{2}-x\right]_{x=4}-\left[\frac{x^{2}}{2}-x\right]_{x=1}\right)
\displaystyle \Rightarrow I=\left(-\left(\frac{1}{2}-1\right)-\left(-0\right)\right)+\left(\left(8-4\right)-\left(\frac{1}{2}-1\right)\right)
\displaystyle \Rightarrow I=\left(-\left(-\frac{1}{2}\right)\right)+\left(4-\left(-\frac{1}{2}\right)\right)
\displaystyle \Rightarrow I=\frac{1}{2}+\frac{9}{2}
\displaystyle \Rightarrow I=5

\displaystyle \textbf{Question 17: }~\int_{1}^{4}\big(\lvert x-1\rvert+\lvert x-2\rvert+\lvert x-4\rvert\big)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{1}^{4}\left(|x-1|+|x-2|+|x-4|\right)\,dx
\displaystyle \Rightarrow I=\int_{1}^{4}|x-1|\,dx+\int_{1}^{4}|x-2|\,dx+\int_{1}^{4}|x-4|\,dx
\displaystyle \text{We know that } |x-1|=\begin{cases}x-1, & 1\le x\le 4\end{cases}
\displaystyle |x-2|=\begin{cases}2-x, & 1\le x\le 2\\ x-2, & 2\le x\le 4\end{cases}
\displaystyle |x-4|=\begin{cases}4-x, & 1\le x\le 4\end{cases}
\displaystyle \Rightarrow I=\int_{1}^{4}(x-1)\,dx+\int_{1}^{2}(2-x)\,dx+\int_{2}^{4}(x-2)\,dx+\int_{1}^{4}(4-x)\,dx
\displaystyle \Rightarrow I=\left[\frac{x^{2}}{2}-x\right]_{1}^{4}+\left[2x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2}}{2}-2x\right]_{2}^{4}+\left[4x-\frac{x^{2}}{2}\right]_{1}^{4}
\displaystyle \Rightarrow I=\left((8-4)-\left(\frac{1}{2}-1\right)\right)+\left((4-2)-\left(2-\frac{1}{2}\right)\right)+\left((8-8)-\left(2-4\right)\right)+\left((16-8)-\left(4-\frac{1}{2}\right)\right)
\displaystyle \Rightarrow I=\left(4+\frac{1}{2}\right)+\left(2-\frac{3}{2}\right)+\left(0+2\right)+\left(8-\frac{7}{2}\right)
\displaystyle \Rightarrow I=\frac{9}{2}+\frac{1}{2}+2+\frac{9}{2}
\displaystyle \Rightarrow I=\frac{23}{2}

\displaystyle \textbf{Question 18: }~\int_{-5}^{0} f(x)\,dx,\ \text{where }f(x)=\lvert x\rvert+\lvert x+2\rvert+\lvert x+5\rvert.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-5}^{0}\left(|x|+|x+2|+|x+5|\right)\,dx
\displaystyle \Rightarrow I=\int_{-5}^{0}|x|\,dx+\int_{-5}^{0}|x+2|\,dx+\int_{-5}^{0}|x+5|\,dx
\displaystyle \text{We know that } |x|=\begin{cases}-x, & -5\le x\le 0\end{cases}
\displaystyle |x+2|=\begin{cases}-(x+2), & -5\le x\le -2\\ x+2, & -2\le x\le 0\end{cases}
\displaystyle |x+5|=\begin{cases}x+5, & -5\le x\le 0\end{cases}
\displaystyle \Rightarrow I=\int_{-5}^{0} (-x)\,dx+\int_{-5}^{-2} (-(x+2))\,dx+\int_{-2}^{0} (x+2)\,dx+\int_{-5}^{0} (x+5)\,dx
\displaystyle \Rightarrow I=\left[-\frac{x^{2}}{2}\right]_{-5}^{0}+\left[-\left(\frac{x^{2}}{2}+2x\right)\right]_{-5}^{-2}+\left[\frac{x^{2}}{2}+2x\right]_{-2}^{0}+\left[\frac{x^{2}}{2}+5x\right]_{-5}^{0}
\displaystyle \Rightarrow I=\left(0-\left(-\frac{25}{2}\right)\right)+\left(-\left(2-4\right)-\left(-\left(\frac{25}{2}-10\right)\right)\right)+\left((0+0)-\left(2-4\right)\right)+\left((0+0)-\left(\frac{25}{2}-25\right)\right)
\displaystyle \Rightarrow I=\frac{25}{2}+\left(2+\frac{5}{2}\right)+\left(0+2\right)+\frac{25}{2}
\displaystyle \Rightarrow I=\frac{25}{2}+\frac{9}{2}+2+\frac{25}{2}
\displaystyle \Rightarrow I=\frac{63}{2}
\displaystyle \textbf{Question 19: }~\int_{0}^{4}\big(\lvert x\rvert+\lvert x-2\rvert+\lvert x-4\rvert\big)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{4}\left(|x|+|x-2|+|x-4|\right)\,dx
\displaystyle \Rightarrow I=\int_{0}^{4}|x|\,dx+\int_{0}^{4}|x-2|\,dx+\int_{0}^{4}|x-4|\,dx
\displaystyle \text{We know that } |x|=\begin{cases}x, & 0\le x\le 4\end{cases}
\displaystyle |x-2|=\begin{cases}2-x, & 0\le x\le 2\\ x-2, & 2\le x\le 4\end{cases}
\displaystyle |x-4|=\begin{cases}4-x, & 0\le x\le 4\end{cases}
\displaystyle \Rightarrow I=\int_{0}^{4} x\,dx+\int_{0}^{2} (2-x)\,dx+\int_{2}^{4} (x-2)\,dx+\int_{0}^{4} (4-x)\,dx
\displaystyle \Rightarrow I=\left[\frac{x^{2}}{2}\right]_{0}^{4}+\left[2x-\frac{x^{2}}{2}\right]_{0}^{2}+\left[\frac{x^{2}}{2}-2x\right]_{2}^{4}+\left[4x-\frac{x^{2}}{2}\right]_{0}^{4}
\displaystyle \Rightarrow I=\left(8-0\right)+\left((4-2)-(0-0)\right)+\left((8-8)-(2-4)\right)+\left((16-8)-(0-0)\right)
\displaystyle \Rightarrow I=8+2+2+8
\displaystyle \Rightarrow I=20

\displaystyle \textbf{Question 20: }~\int_{-1}^{2}\big(\lvert x+1\rvert+\lvert x\rvert+\lvert x-1\rvert\big)\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{We know that}
\displaystyle |x+1|=\begin{cases}x+1, & x\ge -1\\ -(x+1), & x<-1\end{cases}
\displaystyle |x|=\begin{cases}x, & x\ge 0\\ -x, & x<0\end{cases}
\displaystyle |x-1|=\begin{cases}x-1, & x\ge 1\\ -(x-1), & x<1\end{cases}
\displaystyle \text{When } -1\le x\le 0,
\displaystyle |x+1|+|x|+|x-1|=(x+1)+(-x)+(-(x-1))=2-x
\displaystyle \text{When } 0\le x\le 1,
\displaystyle |x+1|+|x|+|x-1|=(x+1)+x+(-(x-1))=x+2
\displaystyle \text{When } 1\le x\le 2,
\displaystyle |x+1|+|x|+|x-1|=(x+1)+x+(x-1)=3x
\displaystyle \therefore I=\int_{-1}^{2}\left(|x+1|+|x|+|x-1|\right)\,dx
\displaystyle \Rightarrow I=\int_{-1}^{0} (2-x)\,dx+\int_{0}^{1} (x+2)\,dx+\int_{1}^{2} 3x\,dx
\displaystyle \Rightarrow I=\left[2x-\frac{x^{2}}{2}\right]_{-1}^{0}+\left[\frac{x^{2}}{2}+2x\right]_{0}^{1}+\left[\frac{3x^{2}}{2}\right]_{1}^{2}
\displaystyle \Rightarrow I=\left(\left(0-0\right)-\left(-2-\frac{1}{2}\right)\right)+\left(\left(\frac{1}{2}+2\right)-0\right)+\left(\frac{3\cdot 4}{2}-\frac{3\cdot 1}{2}\right)
\displaystyle \Rightarrow I=\frac{5}{2}+\frac{5}{2}+\frac{9}{2}
\displaystyle \Rightarrow I=\frac{19}{2}

\displaystyle \textbf{Question 21: }~\int_{-2}^{2}x e^{\lvert x\rvert}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Consider } f(x)=x e^{|x|}
\displaystyle \text{Now, } f(-x)=(-x)e^{|-x|}=-x e^{|x|}=-f(x)
\displaystyle \Rightarrow f(x)\text{ is an odd function.}
\displaystyle \therefore \int_{-2}^{2} x e^{|x|}\,dx=0
\displaystyle \text{In general, } \int_{-a}^{a} f(x)\,dx=\begin{cases}2\int_{0}^{a} f(x)\,dx, & \text{if } f(-x)=f(x)\\ 0, & \text{if } f(-x)=-f(x)\end{cases}

\displaystyle \textbf{Question 22: }~\int_{-\pi/4}^{\pi/2}\sin x\,\lvert \sin x\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x\,|\sin x|\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{4}}^{0} \sin x\,|\sin x|\,dx+\int_{0}^{\frac{\pi}{2}} \sin x\,|\sin x|\,dx
\displaystyle \text{We know that } |\sin x|=\begin{cases}-\sin x, & -\frac{\pi}{4}\le x\le 0\\ \sin x, & 0\le x\le \frac{\pi}{2}\end{cases}
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{4}}^{0} \sin x(-\sin x)\,dx+\int_{0}^{\frac{\pi}{2}} \sin x(\sin x)\,dx
\displaystyle \Rightarrow I=-\int_{-\frac{\pi}{4}}^{0} \sin^{2}x\,dx+\int_{0}^{\frac{\pi}{2}} \sin^{2}x\,dx
\displaystyle \Rightarrow I=-\int_{-\frac{\pi}{4}}^{0} \frac{1-\cos 2x}{2}\,dx+\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2x}{2}\,dx
\displaystyle \Rightarrow I=-\frac{1}{2}\int_{-\frac{\pi}{4}}^{0} dx+\frac{1}{2}\int_{-\frac{\pi}{4}}^{0} \cos 2x\,dx+\frac{1}{2}\int_{0}^{\frac{\pi}{2}} dx-\frac{1}{2}\int_{0}^{\frac{\pi}{2}} \cos 2x\,dx
\displaystyle \Rightarrow I=-\frac{1}{2}\left[x\right]_{-\frac{\pi}{4}}^{0}+\frac{1}{2}\left[\frac{\sin 2x}{2}\right]_{-\frac{\pi}{4}}^{0}+\frac{1}{2}\left[x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\left[\frac{\sin 2x}{2}\right]_{0}^{\frac{\pi}{2}}
\displaystyle \Rightarrow I=-\frac{1}{2}\left(0+\frac{\pi}{4}\right)+\frac{1}{4}\left(0+\sin\frac{\pi}{2}\right)+\frac{1}{2}\left(\frac{\pi}{2}-0\right)-\frac{1}{4}\left(\sin\pi-0\right)
\displaystyle \Rightarrow I=-\frac{\pi}{8}+\frac{1}{4}+\frac{\pi}{4}-0
\displaystyle \Rightarrow I=\frac{\pi}{8}+\frac{1}{4}

\displaystyle \textbf{Question 23: }~\int_{0}^{\pi}\cos x\,\lvert \cos x\rvert\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Consider } f(x)=\cos x\,|\cos x|
\displaystyle \text{Now, } f(\pi-x)=\cos(\pi-x)\,|\cos(\pi-x)|
\displaystyle \Rightarrow f(\pi-x)=(-\cos x)\,|-\cos x|=-\cos x\,|\cos x|=-f(x)
\displaystyle \therefore \int_{0}^{\pi} \cos x\,|\cos x|\,dx=0
\displaystyle \text{In general, } \int_{0}^{2a} f(x)\,dx=\begin{cases}2\int_{0}^{a} f(x)\,dx, & \text{if } f(2a-x)=f(x)\\ 0, & \text{if } f(2a-x)=-f(x)\end{cases}

\displaystyle \textbf{Question 24: }~\int_{-\pi/2}^{\pi/2}\big(2\sin\lvert x\rvert+\cos \lvert x\rvert\big)\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Consider } f(x)=2\sin|x|+\cos|x|
\displaystyle \text{Now, } f(-x)=2\sin|-x|+\cos|-x|=2\sin|x|+\cos|x|=f(x)
\displaystyle \Rightarrow f(x)\text{ is an even function.}
\displaystyle \therefore \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(2\sin|x|+\cos|x|\right)\,dx=2\int_{0}^{\frac{\pi}{2}} \left(2\sin|x|+\cos|x|\right)\,dx
\displaystyle \Rightarrow \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(2\sin|x|+\cos|x|\right)\,dx=2\int_{0}^{\frac{\pi}{2}} \left(2\sin x+\cos x\right)\,dx
\displaystyle \Rightarrow \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(2\sin|x|+\cos|x|\right)\,dx=4\int_{0}^{\frac{\pi}{2}} \sin x\,dx+2\int_{0}^{\frac{\pi}{2}} \cos x\,dx
\displaystyle \Rightarrow \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(2\sin|x|+\cos|x|\right)\,dx=4\left[-\cos x\right]_{0}^{\frac{\pi}{2}}+2\left[\sin x\right]_{0}^{\frac{\pi}{2}}
\displaystyle \Rightarrow \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(2\sin|x|+\cos|x|\right)\,dx=-4\left(\cos\frac{\pi}{2}-\cos 0\right)+2\left(\sin\frac{\pi}{2}-\sin 0\right)
\displaystyle \Rightarrow \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(2\sin|x|+\cos|x|\right)\,dx=-4(0-1)+2(1-0)
\displaystyle \Rightarrow \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(2\sin|x|+\cos|x|\right)\,dx=4+2
\displaystyle \Rightarrow \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(2\sin|x|+\cos|x|\right)\,dx=6

\displaystyle \textbf{Question 25: }~\int_{-\pi/2}^{\pi}\sin^{-1}(\sin x)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{-\frac{\pi}{2}}^{\pi} \sin^{-1}(\sin x)\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{-1}(\sin x)\,dx+\int_{\frac{\pi}{2}}^{\pi} \sin^{-1}(\sin x)\,dx
\displaystyle \text{For } -\frac{\pi}{2}\le x\le \frac{\pi}{2},\ \sin^{-1}(\sin x)=x
\displaystyle \text{For } \frac{\pi}{2}\le x\le \pi,\ \sin^{-1}(\sin x)=\pi-x
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x\,dx+\int_{\frac{\pi}{2}}^{\pi} (\pi-x)\,dx
\displaystyle \Rightarrow I=\left[\frac{x^{2}}{2}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\left[\pi x-\frac{x^{2}}{2}\right]_{\frac{\pi}{2}}^{\pi}
\displaystyle \Rightarrow I=\left(\frac{\pi^{2}}{8}-\frac{\pi^{2}}{8}\right)+\left(\pi^{2}-\frac{\pi^{2}}{2}-\left(\frac{\pi^{2}}{2}-\frac{\pi^{2}}{8}\right)\right)
\displaystyle \Rightarrow I=0+\frac{\pi^{2}}{8}
\displaystyle \Rightarrow I=\frac{\pi^{2}}{8}

\displaystyle \text{Question 26: }~\int_{-\pi/2}^{\pi/2}\frac{-\pi/2}{\sqrt{\cos x\,\sin^2 x}}\,dx.
\displaystyle \text{Answer:}

\displaystyle \textbf{Question 27: }~\int_{0}^{2}2x\,[x]\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{2} 2x[x]\,dx
\displaystyle \Rightarrow I=\int_{0}^{1} 2x[x]\,dx+\int_{1}^{2} 2x[x]\,dx
\displaystyle \text{We know that } [x]=\begin{cases}0, & 0\le x<1\\ 1, & 1\le x<2\end{cases}
\displaystyle \Rightarrow I=\int_{0}^{1} 2x\cdot 0\,dx+\int_{1}^{2} 2x\cdot 1\,dx
\displaystyle \Rightarrow I=0+2\int_{1}^{2} x\,dx
\displaystyle \Rightarrow I=2\left[\frac{x^{2}}{2}\right]_{1}^{2}
\displaystyle \Rightarrow I=2\left(2-\frac{1}{2}\right)
\displaystyle \Rightarrow I=3

\displaystyle \textbf{Question 28: }~\int_{0}^{2\pi}\cos^{-1}(\cos x)\,dx.
\displaystyle \text{Answer:}
\displaystyle I=\int_{0}^{2} 2x[x]\,dx
\displaystyle \Rightarrow I=\int_{0}^{1} 2x[x]\,dx+\int_{1}^{2} 2x[x]\,dx
\displaystyle \text{We know that } [x]=\begin{cases}0, & 0\le x<1\\ 1, & 1\le x<2\end{cases}
\displaystyle \Rightarrow I=\int_{0}^{1} 2x\cdot 0\,dx+\int_{1}^{2} 2x\cdot 1\,dx
\displaystyle \Rightarrow I=0+2\int_{1}^{2} x\,dx
\displaystyle \Rightarrow I=2\left[\frac{x^{2}}{2}\right]_{1}^{2}
\displaystyle \Rightarrow I=2\left(2-\frac{1}{2}\right)
\displaystyle \Rightarrow I=3

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