Find the area under the curve y =f( x) on [a,b] given f(x)=tan(3x) where a=0 b=pi/12

Answer:The area under the curve y=f(x) on [a,b] is [tex]\frac{1}{6}\ln(2)[/tex] square units.Step-by-step explanation:The given function is[tex]f(x)=\tan(3x)[/tex]where a=0 and b=pi/12.The area under the curve y=f(x) on [a,b] is defined as[tex]Area=\int_{a}^{b}f(x)dx[/tex][tex]Area=\int_{0}^{\frac{\pi}{12}}\tan (3x)dx[/tex][tex]Area=\int_{0}^{\frac{\pi}{12}}\frac{\sin (3x)}{\cos (3x)}dx[/tex]Substitute cos (3x)=t, so[tex]-3\sin (3x)dx=dt[/tex][tex]\sin (3x)dx=-\frac{1}{3}dt[/tex][tex]a=\cos (3(0))=1[/tex][tex]b=\cos (3(\frac{\pi}{12}))=\frac{1}{\sqrt{2}}[/tex][tex]Area=-\frac{1}{3}\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{t}dt[/tex][tex]Area=-\frac{1}{3}[\ln t]_{1}^{\frac{1}{\sqrt{2}}[/tex][tex]Area=-\frac{1}{3}(\ln \frac{1}{\sqrt{2}}-\ln (1))[/tex][tex]Area=-\frac{1}{3}(\ln 1-\ln \sqrt{2}-0)[/tex][tex]Area=-\frac{1}{3}(-\ln 2^{\frac{1}{2}})[/tex][tex]Area=-\frac{1}{3}(-\frac{1}{2}\ln 2)[/tex][tex]Area=-\frac{1}{6}\ln 2[/tex]Therefore the area under the curve y=f(x) on [a,b] is [tex]\frac{1}{6}\ln(2)[/tex] square units.