What is the integral from 0 to 1 of [3x+cos(3x-3)] ?

The integral from 0 to 1 of [3x+cos(3x-3)] is 3/2+1/3 sin(3)

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integral from 0 to 1 of [3x+cos(3x-3)]

\int_{0}^{1} [3 x + cos (3 x - 3)] d x

\frac{3}{2} + \frac{1}{3} sin (3)

Decimal

1.54704 \dots

Solution steps

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

\int_{0}^{1} 3 x d x = \frac{3}{2}

\int_{0}^{1} cos (3 x - 3) d x = \frac{1}{3} sin (3)

= \frac{3}{2} + \frac{1}{3} sin (3)

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What is the integral from 0 to 1 of [3x+cos(3x-3)] ?
The integral from 0 to 1 of [3x+cos(3x-3)] is 3/2+1/3 sin(3)