What is the integral from 0 to 2pi of 6x^2sin(x/4) ?

The integral from 0 to 2pi of 6x^2sin(x/4) is 384(pi-2)

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integral from 0 to 2pi of 6x^2sin(x/4)

\int_{0}^{2 π} 6 x^{2} sin (\frac{x}{4}) d x

384 (π - 2)

Decimal Notation

438.37157 \dots

Solution steps

\int_{0}^{2 π} 6 x^{2} sin (\frac{x}{4}) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{2 π} x^{2} sin (\frac{x}{4}) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} 64 u^{2} sin (u) d u

= 6 \cdot \int_{0}^{\frac{π}{2}} 64 u^{2} sin (u) d u

Take the constant out:

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

= 6 \cdot 64 \cdot \int_{0}^{\frac{π}{2}} u^{2} sin (u) d u

Apply Integration By Parts

: [- u^{2} cos (u) - \int - 2 u cos (u) d u]_{0}^{\frac{π}{2}}

= 6 \cdot 64 [- u^{2} cos (u) - \int - 2 u cos (u) d u]_{0}^{\frac{π}{2}}

\int - 2 u cos (u) d u = - 2 (u sin (u) + cos (u))

= 6 \cdot 64 [- u^{2} cos (u) - (- 2 (u sin (u) + cos (u)))]_{0}^{\frac{π}{2}}

Simplify

= 384 [- u^{2} cos (u) + 2 (u sin (u) + cos (u))]_{0}^{\frac{π}{2}}

Compute the boundaries:

π - 2

= 384 (π - 2)

\frac{\partial}{\partial x} (e^{x yz} tan (x + 2 y + z))

x sin (\frac{2}{x})

y = 4 x^{3} + x, x = 2, x = - 1, x

x \to - 2 lim (\frac{x - 1}{x + 2})

\frac{1}{y ^{2}} - \frac{3}{y ^{4}}

What is the integral from 0 to 2pi of 6x^2sin(x/4) ?
The integral from 0 to 2pi of 6x^2sin(x/4) is 384(pi-2)