What is area 8cos(2x),8-8cos(2x),0, pi/2 ?

The answer to area 8cos(2x),8-8cos(2x),0, pi/2 is -(8pi)/3+4pi+8sqrt(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area 8cos(2x),8-8cos(2x),0, pi/2

area

8 cos (2 x), 8 - 8 cos (2 x), 0, \frac{π}{2}

- \frac{8 π}{3} + 4 π + 83

Decimal Notation

18.04519 \dots

Solution steps

Apply the area formula:

\int_{0}^{\frac{π}{2}} ∣ 8 cos (2 x) - (8 - 8 cos (2 x)) ∣ d x

Solve

\int_{0}^{\frac{π}{2}} ∣ 8 cos (2 x) - (8 - 8 cos (2 x)) ∣ d x : - \frac{8 π}{3} + 4 π + 83

The area is:

= - \frac{8 π}{3} + 4 π + 83

x \to 0 + lim (x - \frac{1}{x ^{3}})

\frac{\partial}{\partial x} (ln (1 + x^{2} + e^{y} + z))

y = sec (x), (\frac{π}{6}, \frac{2 3}{3})

2 x y^{^{'}} - 8 y = x^{- 4}

- \frac{1}{2 1 - x}

What is area 8cos(2x),8-8cos(2x),0, pi/2 ?
The answer to area 8cos(2x),8-8cos(2x),0, pi/2 is -(8pi)/3+4pi+8sqrt(3)