What is the integral from 0 to 1 of sin(pi)t ?

The integral from 0 to 1 of sin(pi)t is 0

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integral from 0 to 1 of sin(pi)t

\int_{0}^{1} sin (π) t d t

0

Solution steps

\int_{0}^{1} sin (π) t d t

sin (π) t = 0

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

Integral of a constant:

\int a d x = a x

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

Simplify

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

Compute the boundaries:

0

= 0

\int_{5}^{6} x (x - 5)^{2} d x

\int_{0}^{1} x 4^{x} d x

\int_{0}^{1} (x - x^{3})^{2} d x

\int_{3}^{12} x - 3 d x

\int_{0}^{2} \frac{24}{x ^{2} + 4} d x

What is the integral from 0 to 1 of sin(pi)t ?
The integral from 0 to 1 of sin(pi)t is 0