What is the integral from 0 to 1 of sin(3pit) ?

The integral from 0 to 1 of sin(3pit) is 2/(3pi)

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integral from 0 to 1 of sin(3pit)

\int_{0}^{1} sin (3 π t) d t

\frac{2}{3 π}

Solution steps

\int_{0}^{1} sin (3 π t) d t

Apply u-substitution

: \int_{0}^{3 π} sin (u) \frac{1}{3 π} d u

= \int_{0}^{3 π} sin (u) \frac{1}{3 π} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3 π} \cdot \int_{0}^{3 π} sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{3 π} [- cos (u)]_{0}^{3 π}

Compute the boundaries:

2

= \frac{1}{3 π} \cdot 2

Simplify

= \frac{2}{3 π}

\int_{2}^{4} 3 x^{2} ln (2 x) d x

\int_{1}^{e^{x}} 4 ln (x) d x

\int_{0}^{1} 2 x y d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} (- cos (x)) d x

\int_{0}^{x} \frac{3}{x} d x

What is the integral from 0 to 1 of sin(3pit) ?
The integral from 0 to 1 of sin(3pit) is 2/(3pi)