What is the integral from 0 to 1 of x^2cos(3x) ?

The integral from 0 to 1 of x^2cos(3x) is 1/27 (7sin(3)+6cos(3))

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integral from 0 to 1 of x^2cos(3x)

\int_{0}^{1} x^{2} cos (3 x) d x

\frac{1}{27} (7 sin (3) + 6 cos (3))

Decimal Notation

- 0.18341 \dots

Solution steps

\int_{0}^{1} x^{2} cos (3 x) d x

Apply u-substitution

: \int_{0}^{3} \frac{u ^{2} cos ( u )}{27} d u

= \int_{0}^{3} \frac{u ^{2} cos ( u )}{27} d u

Take the constant out:

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

= \frac{1}{27} \cdot \int_{0}^{3} u^{2} cos (u) d u

Apply Integration By Parts

: [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{3}

= \frac{1}{27} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{3}

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

= \frac{1}{27} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{3}

Compute the boundaries:

7 sin (3) + 6 cos (3)

= \frac{1}{27} (7 sin (3) + 6 cos (3))

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What is the integral from 0 to 1 of x^2cos(3x) ?
The integral from 0 to 1 of x^2cos(3x) is 1/27 (7sin(3)+6cos(3))