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

Answer:

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

Step-by-step 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

$Area=\int_{a}^{b}f(x)dx$

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

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

Substitute cos (3x)=t, so

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

$\sin (3x)dx=-\frac{1}{3}dt$

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

$b=\cos (3(\frac{\pi}{12}))=\frac{1}{\sqrt{2}}$

$Area=-\frac{1}{3}\int_{1}^{\frac{1}{\sqrt{2}}}\frac{1}{t}dt$

$Area=-\frac{1}{3}[\ln t]_{1}^{\frac{1}{\sqrt{2}}$

$Area=-\frac{1}{3}(\ln \frac{1}{\sqrt{2}}-\ln (1))$

$Area=-\frac{1}{3}(\ln 1-\ln \sqrt{2}-0)$

$Area=-\frac{1}{3}(-\ln 2^{\frac{1}{2}})$

$Area=-\frac{1}{3}(-\frac{1}{2}\ln 2)$

$Area=-\frac{1}{6}\ln 2$

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