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

The integral from 0 to 2 of (2x-cos(4x)) is 4-1/4 sin(8)

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integral from 0 to 2 of (2x-cos(4x))

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

4 - \frac{1}{4} sin (8)

Decimal Notation

3.75266 \dots

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{2} 2 x d x = 4

\int_{0}^{2} cos (4 x) d x = \frac{1}{4} sin (8)

= 4 - \frac{1}{4} sin (8)

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What is the integral from 0 to 2 of (2x-cos(4x)) ?
The integral from 0 to 2 of (2x-cos(4x)) is 4-1/4 sin(8)