What is the integral from 0 to 2pi of t^2sin(2) ?

The integral from 0 to 2pi of t^2sin(2) is sin(2)(8pi^3)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2pi of t^2sin(2)

\int_{0}^{2 π} t^{2} sin (2) d t

sin (2) \frac{8 π ^{3}}{3}

Decimal Notation

75.18380 \dots

Solution steps

\int_{0}^{2 π} t^{2} sin (2) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= sin (2) \cdot \int_{0}^{2 π} t^{2} d t

Apply the Power Rule

: [\frac{t ^{3}}{3}]_{0}^{2 π}

= sin (2) [\frac{t ^{3}}{3}]_{0}^{2 π}

Compute the boundaries:

\frac{8 π ^{3}}{3}

= sin (2) \frac{8 π ^{3}}{3}

\frac{\partial}{\partial v} (sin (u) sin (v))

x \to - 2 lim (2 x^{2} + 5 x + 3)

y = \frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}}

f (x) = (3 + 2 x) e^{- 3 x}

\int_{0}^{2} (3 x^{2} - 6 x + 1) d x

What is the integral from 0 to 2pi of t^2sin(2) ?
The integral from 0 to 2pi of t^2sin(2) is sin(2)(8pi^3)/3