Question

How do you do this question?

Answer 1

7√3 + 7π/6 ≈ 15.790

Explanation:

First, find the intersection:

7 cos(2x) = 7 − 7 cos(2x)

14 cos(2x) = 7

cos(2x) = 1/2

2x = π/3

x = π/6

From the graph:

On 0 ≤ x ≤ π/6, 7 cos(2x) > 7 − 7 cos(2x).

On π/6 ≤ x ≤ π/2, 7 − 7 cos(2x) > 7 cos(2x).

So the integral is:

`\int\limits^(\pi)/(6) _0 {[7cos(2x)-(7-7cos(2x))]} \, dx + \int\limits^(\pi)/(2) _(\pi)/(6) {[(7-7cos(2x))-7cos(2x)]} \, dx`

`\int\limits^(\pi)/(6) _0 {[7cos(2x)-7+7cos(2x))]} \, dx + \int\limits^(\pi)/(2) _(\pi)/(6) {[7-7cos(2x)-7cos(2x)]} \, dx`

`\int\limits^(\pi)/(6) _0 {[14cos(2x)-7]} \, dx + \int\limits^(\pi)/(2) _(\pi)/(6) {[7-14cos(2x)]} \, dx`

`[7sin(2x)-7x]|^(\pi)/(6) _0 + [7x-7sin(2x)]|^(\pi)/(2) _(\pi)/(6)`

`[7sin((\pi )/(3) )-(7\pi )/(6) ] - [7sin(0)-0 ] + [(7\pi )/(2) -7sin(\pi )] - [(7\pi )/(6) -7sin((\pi )/(3) )]`

`7sin((\pi )/(3) )-(7\pi )/(6) + (7\pi )/(2) -7sin(\pi ) - (7\pi )/(6) +7sin((\pi )/(3) )`

`14sin((\pi )/(3) )-(7\pi )/(3) + (7\pi )/(2)`

`7√(3) + (7\pi )/(6)`