What is the integral from 0 to 4 of (xcos(x^2))/2 ?

The integral from 0 to 4 of (xcos(x^2))/2 is 1/4 sin(16)

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integral from 0 to 4 of (xcos(x^2))/2

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

\frac{1}{4} sin (16)

Decimal

- 0.07197 \dots

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{4} x cos (x^{2}) d x

Apply u-substitution

: \int_{0}^{16} \frac{cos ( u )}{2} d u

= \frac{1}{2} \cdot \int_{0}^{16} \frac{cos ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{16} cos (u) d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} [sin (u)]_{0}^{16}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [sin (u)]_{0}^{16} : \frac{1}{4} [sin (u)]_{0}^{16}

= \frac{1}{4} [sin (u)]_{0}^{16}

Compute the boundaries:

sin (16)

= \frac{1}{4} sin (16)

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What is the integral from 0 to 4 of (xcos(x^2))/2 ?
The integral from 0 to 4 of (xcos(x^2))/2 is 1/4 sin(16)