What is the integral from 0 to 2pi of-16 ?

The integral from 0 to 2pi of-16 is -32pi

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integral from 0 to 2pi of-16

\int_{0}^{2 π} - 16 d t

- 32 π

Decimal

- 100.53096 \dots

Solution steps

\int_{0}^{2 π} - 16 d t

Integral of a constant:

\int a d x = a x

= [(- 16) t]_{0}^{2 π}

Simplify

= [- 16 t]_{0}^{2 π}

Compute the boundaries:

- 32 π

= - 32 π

\int_{- π}^{π} cos^{2} (m x) d x

\int_{0}^{8} e^{x} d x

\int_{- 1}^{3} e^{x} - 1 d x

\int_{0}^{x^{2}} e^{t} d t

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

What is the integral from 0 to 2pi of-16 ?
The integral from 0 to 2pi of-16 is -32pi