What is the integral from 0 to pi/2 of 3xcos(2x) ?

The integral from 0 to pi/2 of 3xcos(2x) is -3/2

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integral from 0 to pi/2 of 3xcos(2x)

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

- \frac{3}{2}

Decimal Notation

- 1.5

Solution steps

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

Take the constant out:

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

= 3 \cdot \int_{0}^{\frac{π}{2}} x cos (2 x) d x

Apply u-substitution

: \int_{0}^{π} \frac{u cos ( u )}{4} d u

= 3 \cdot \int_{0}^{π} \frac{u cos ( u )}{4} d u

Take the constant out:

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

= 3 \cdot \frac{1}{4} \cdot \int_{0}^{π} u cos (u) d u

Apply Integration By Parts

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

= 3 \cdot \frac{1}{4} [u sin (u) - \int sin (u) d u]_{0}^{π}

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

= 3 \cdot \frac{1}{4} [u sin (u) - (- cos (u))]_{0}^{π}

Simplify

3 \cdot \frac{1}{4} [u sin (u) - (- cos (u))]_{0}^{π} : \frac{3}{4} [u sin (u) + cos (u)]_{0}^{π}

= \frac{3}{4} [u sin (u) + cos (u)]_{0}^{π}

Compute the boundaries:

- 2

= \frac{3}{4} (- 2)

Simplify

= - \frac{3}{2}

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\int_{- 2}^{\infty} e^{- x} d x

What is the integral from 0 to pi/2 of 3xcos(2x) ?
The integral from 0 to pi/2 of 3xcos(2x) is -3/2