Physics, asked by jesvipatel, 1 month ago

Can anyone please solve this? ​

Attachments:

Answers

Answered by Anonymous
12

\maltese\:\underline{\underline{\textsf{Answer :}}}\maltese

\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}

\longrightarrow \: \displaystyle \rm y =  \int\limits_{- 4}^{ - 1}  \dfrac{\pi}{2}  \, d \theta  \\

\longrightarrow \: \displaystyle \rm y = \dfrac{\pi}{2}   \int\limits_{- 4}^{ - 1}    \, d \theta  \\

\longrightarrow \: \displaystyle \rm y = \dfrac{\pi}{2}\Bigg[ \theta\Bigg]_{ - 4}^{ - 1}  \\

\longrightarrow \: \displaystyle \rm y = \dfrac{\pi}{2}\Bigg[  - 1 - ( - 4)\Bigg]  \\

\longrightarrow \: \displaystyle \rm y = \dfrac{\pi}{2}\Bigg[  - 1  + 4\Bigg]  \\

\longrightarrow \: \displaystyle \rm y = \dfrac{\pi}{2}\bigg[ 3\bigg]  \\

\longrightarrow \: \displaystyle  \underline{ \underline{\rm y = \dfrac{3\pi}{2}}}  \\

\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}

\longrightarrow \: \displaystyle \rm y =  \int\limits_{ \sqrt{2} }^{ 5 \sqrt{2} }  r  \, dr \\

\longrightarrow \: \displaystyle \rm y = \Bigg[   \dfrac{ {r}^{2} }{2}  \Bigg]_{ \sqrt{2} }^{  5\sqrt{2} }  \\

\longrightarrow \: \displaystyle \rm y = \dfrac{ {(5  \sqrt{2}  )}^{2} }{2}  - \dfrac{ {(\sqrt{2}  )}^{2} }{2}   \\

\longrightarrow \: \displaystyle \rm y = \dfrac{ 25 \times 2 }{2}  - \dfrac{2}{2}   \\

\longrightarrow \: \displaystyle \rm y = \dfrac{ 50 }{2}  - \dfrac{2 }{2}   \\

\longrightarrow \: \displaystyle \rm y = \dfrac{ 50  - 2}{2}   \\

\longrightarrow \: \displaystyle \rm y = \dfrac{ 48}{2}   \\

\longrightarrow \: \displaystyle  \underline{ \underline{\rm y = 24}}  \\

\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}

\longrightarrow \: \displaystyle \rm y =  \int\limits_{ 0 }^{1 }   {e}^{x}  \, dx \\

\longrightarrow \: \displaystyle \rm y = \Bigg[  {e}^{x}  \Bigg]_{ 0 }^{ 1 }  \\

\longrightarrow \: \displaystyle \rm y ={e}^{1}   - {e}^{0}   \\

\longrightarrow \: \displaystyle \rm y =e- 1  \\

\longrightarrow \: \displaystyle \rm y =2.718- 1  \\

\longrightarrow \: \displaystyle  \underline{ \underline{\rm y  = 1.718}}  \\

\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.8}}\end{picture}


Anonymous: Awesome!!
Anonymous: Thanks <3
Similar questions