What is the integral from 0 to 2pi of 5x^2sin(1/4 x) ?

The integral from 0 to 2pi of 5x^2sin(1/4 x) is 320(pi-2)

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integral from 0 to 2pi of 5x^2sin(1/4 x)

\int_{0}^{2 π} 5 x^{2} sin (\frac{1}{4} x) d x

320 (π - 2)

Decimal Notation

365.30964 \dots

Solution steps

\int_{0}^{2 π} 5 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

= 5 \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

= 5 \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

= 5 \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}}

= 5 \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))

= 5 \cdot 64 [- u^{2} cos (u) - (- 2 (u sin (u) + cos (u)))]_{0}^{\frac{π}{2}}

Simplify

= 320 [- u^{2} cos (u) + 2 (u sin (u) + cos (u))]_{0}^{\frac{π}{2}}

Compute the boundaries:

π - 2

= 320 (π - 2)

\frac{d}{d x} (\frac{1}{3} x x^{2} + 5)

\int \frac{1}{- 4 t ^{2} + 12 t + 7} d t

\int_{0}^{π} sin (x^{2}) d x

\int cos (t) + sin^{3} (t) d t

y^{^{'}} + \frac{6}{x ^{2}} y = x^{10} e^{\frac{6}{x}}

What is the integral from 0 to 2pi of 5x^2sin(1/4 x) ?
The integral from 0 to 2pi of 5x^2sin(1/4 x) is 320(pi-2)