The area of the bounded by the curve y = tan x, x-axis and the lines $x = \frac{π}{3} and x = \frac{π}{3} is__________________.$

Open in App

Solution

To find: area of the region bounded by the curve y = tan x, x-axis and the lines $x = - \frac{π}{3} and x = \frac{π}{3}$

The required area of the region = $\int_{- \frac{π}{3}}^{\frac{π}{3}} y d x$
Thus,
$Required area = \int_{- \frac{π}{3}}^{\frac{π}{3}} y d x = \int_{- \frac{π}{3}}^{\frac{π}{3}} \tan x d x = |\int_{- \frac{π}{3}}^{0} \tan x d x| + \int_{0}^{\frac{π}{3}} \tan x d x = |{(\sec^{2} x)}_{- \frac{π}{3}}^{0}| + {(\sec^{2} x)}_{0}^{\frac{π}{3}} = |(\sec^{2} 0) - (\sec^{2} (- \frac{π}{3}))| + ((\sec^{2} \frac{π}{3}) - (\sec^{2} 0)) = |(1) - (4)| + ((4) - (1)) = |- 3| + 3 = 3 + 3 = 6 Thus, Area = 6 sq . units$

Hence, the area of the region bounded by the curve y = tan x, x-axis and the lines $x = - \frac{π}{3} and x = \frac{π}{3}$ is $6 sq . units .$

Suggest Corrections

Similar questions

y = \frac{1}{1 + \sqrt{tan x}}

x

x = \frac{π}{6}

x = \frac{π}{3}

y = x^{2}

x

x = 1

y = tan x

x = \frac{π}{4}

x

y = s i n (\frac{x}{3})

x = 0

x = 3 π

Join BYJU'S Learning Program