Question

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

Answer 1

The area under the curve y=f(x) on [a,b] is
`(1)/(6)\ln(2)` square units.

Explanation:

The given function is

`f(x)=\tan(3x)`

where a=0 and b=pi/12.

The area under the curve y=f(x) on [a,b] is defined as

`Area=\int_(a)^(b)f(x)dx`

`Area=\int_(0)^{(\pi)/(12)}\tan (3x)dx`

`Area=\int_(0)^{(\pi)/(12)}(\sin (3x))/(\cos (3x))dx`

Substitute cos (3x)=t, so

`-3\sin (3x)dx=dt`

`\sin (3x)dx=-(1)/(3)dt`

`a=\cos (3(0))=1`

`b=\cos (3((\pi)/(12)))=(1)/(√(2))`

`Area=-(1)/(3)\int_(1)^{(1)/(√(2))}(1)/(t)dt`

`Area=-(1)/(3)[\ln t]_(1)^{(1)/(√(2))`

`Area=-(1)/(3)(\ln (1)/(√(2))-\ln (1))`

`Area=-(1)/(3)(\ln 1-\ln √(2)-0)`

`Area=-(1)/(3)(-\ln 2^{(1)/(2)})`

`Area=-(1)/(3)(-(1)/(2)\ln 2)`

`Area=-(1)/(6)\ln 2`

Therefore the area under the curve y=f(x) on [a,b] is
`(1)/(6)\ln(2)` square units.