What is the integral from 0 to 2pi of 4x^2sin(1/4 x) ?

The integral from 0 to 2pi of 4x^2sin(1/4 x) is 256(pi-2)

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integral from 0 to 2pi of 4x^2sin(1/4 x)

\int_{0}^{2 π} 4 x^{2} sin (\frac{1}{4} x) d x

256 (π - 2)

Decimal Notation

292.24771 \dots

Solution steps

\int_{0}^{2 π} 4 x^{2} sin (\frac{1}{4} x) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{2 π} x^{2} sin (\frac{1}{4} x) d x

Apply u-substitution

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

= 4 \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

= 4 \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}}

= 4 \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))

= 4 \cdot 64 [- u^{2} cos (u) - (- 2 (u sin (u) + cos (u)))]_{0}^{\frac{π}{2}}

Simplify

= 256 [- u^{2} cos (u) + 2 (u sin (u) + cos (u))]_{0}^{\frac{π}{2}}

Compute the boundaries:

π - 2

= 256 (π - 2)

x \to - 1 lim (3 x^{4} - 2 x^{3} + 4 x)

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

\frac{1}{2 s ^{3} + 10 s ^{2} + 12 s}

y = x^{4} cos (4 x^{2})

y = 17 x, y = x^{2}

What is the integral from 0 to 2pi of 4x^2sin(1/4 x) ?
The integral from 0 to 2pi of 4x^2sin(1/4 x) is 256(pi-2)