What is the integral from 0 to 10 of x^2cos(4x) ?

The integral from 0 to 10 of x^2cos(4x) is 1/64 (1598sin(40)+80cos(40))

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integral from 0 to 10 of x^2cos(4x)

\int_{0}^{10} x^{2} cos (4 x) d x

\frac{1}{64} (1598 sin (40) + 80 cos (40))

Solution steps

\int_{0}^{10} x^{2} cos (4 x) d x

Apply u-substitution

: \int_{0}^{40} \frac{u ^{2} cos ( u )}{64} d u

= \int_{0}^{40} \frac{u ^{2} cos ( u )}{64} d u

Take the constant out:

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

= \frac{1}{64} \cdot \int_{0}^{40} u^{2} cos (u) d u

Apply Integration By Parts

: [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{40}

= \frac{1}{64} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{40}

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

= \frac{1}{64} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{40}

Compute the boundaries:

1598 sin (40) + 80 cos (40)

= \frac{1}{64} (1598 sin (40) + 80 cos (40))

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What is the integral from 0 to 10 of x^2cos(4x) ?
The integral from 0 to 10 of x^2cos(4x) is 1/64 (1598sin(40)+80cos(40))