What is the integral from 0 to 1 of [2x]sin(pi)x ?

The integral from 0 to 1 of [2x]sin(pi)x is 0

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integral from 0 to 1 of [2x]sin(pi)x

\int_{0}^{1} [2 x] sin (π) x d x

0

Solution steps

\int_{0}^{1} [2 x] sin (π) x d x

[2 x] sin (π) x = 0

= \int_{0}^{1} 0 d x

Integral of a constant:

\int a d x = a x

= [0 \cdot x]_{0}^{1}

Simplify

= [0]_{0}^{1}

Compute the boundaries:

0

= 0

\int_{- 2}^{2} \frac{1}{x ^{2} + 1} d x

\int_{0}^{6} 2 x^{2} + 3 x - 1 d x

\int_{- 1}^{2} (5 - x^{2}) d x

\int_{0}^{3} e^{x + 3 y} d x

\int_{0}^{2 π} cos (2 x) sin (5 x) d x

What is the integral from 0 to 1 of [2x]sin(pi)x ?
The integral from 0 to 1 of [2x]sin(pi)x is 0