What is the integral from 0 to 3pi of 2x^2sin(1/6 x) ?

The integral from 0 to 3pi of 2x^2sin(1/6 x) is 432(pi-2)

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integral from 0 to 3pi of 2x^2sin(1/6 x)

\int_{0}^{3 π} 2 x^{2} sin (\frac{1}{6} x) d x

432 (π - 2)

Solution steps

\int_{0}^{3 π} 2 x^{2} sin (\frac{1}{6} 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}^{3 π} x^{2} sin (\frac{1}{6} x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} 216 u^{2} sin (u) d u

= 2 \cdot \int_{0}^{\frac{π}{2}} 216 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 216 \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 216 [- 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 216 [- u^{2} cos (u) - (- 2 (u sin (u) + cos (u)))]_{0}^{\frac{π}{2}}

Simplify

= 432 [- u^{2} cos (u) + 2 (u sin (u) + cos (u))]_{0}^{\frac{π}{2}}

Compute the boundaries:

π - 2

= 432 (π - 2)

\frac{d}{d x} (\frac{e ^{x} + 1}{x ^{2}})

\frac{d}{d x} (6 x^{4} - 2 x^{3} - 12 x^{2} + 3 x + 3)

\frac{d ^{2}}{d x ^{2}} (\frac{4}{5 t ^{6}})

e^{2 - x}

y^{^{'}} = 23 x y

What is the integral from 0 to 3pi of 2x^2sin(1/6 x) ?
The integral from 0 to 3pi of 2x^2sin(1/6 x) is 432(pi-2)