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Mathematics
15.04.2020 06:48
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Find the area under the curve y =f( x) on [a,b] given f(x)=tan(3x) where a=0 b=pi/12

Find the area under the curve y =f( x) on [a,b] given f(x)=tan(3x) where a=0 b=pi/12
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bre2795
bre2795
5,0(12 marks)

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.

beanokelley
beanokelley
4,4(31 marks)
If am right it’s a hope it helps

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