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

The integral from 0 to 2pi of 2x^2sin(1/4 x) is 128(pi-2)

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

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

128 (π - 2)

Solution steps

\int_{0}^{2 π} 2 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

= 2 \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

= 2 \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

= 2 \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}}

= 2 \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))

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

Simplify

= 128 [- u^{2} cos (u) + 2 (u sin (u) + cos (u))]_{0}^{\frac{π}{2}}

Compute the boundaries:

π - 2

= 128 (π - 2)

x \to 0 lim (\frac{1}{x} \cdot ln (x + 1))

\int 6 x yz + e^{x} d y

\frac{\partial}{\partial y} (4 x + 8 y - 88)

x \to 0 + lim ((x) ln (x^{4}))

\frac{40}{1 + 25 p + 100 p ^{2}}

What is the integral from 0 to 2pi of 2x^2sin(1/4 x) ?
The integral from 0 to 2pi of 2x^2sin(1/4 x) is 128(pi-2)