What is the integral from 2 to 3 of 3sin(pi/3 x) ?

The integral from 2 to 3 of 3sin(pi/3 x) is 9/(2pi)

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integral from 2 to 3 of 3sin(pi/3 x)

\int_{2}^{3} 3 sin (\frac{π}{3} x) d x

\frac{9}{2 π}

Decimal

1.43239 \dots

Solution steps

\int_{2}^{3} 3 sin (\frac{π}{3} x) d x

Take the constant out:

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

= 3 \cdot \int_{2}^{3} sin (\frac{π}{3} x) d x

Apply u-substitution

= 3 \cdot \int_{\frac{2 π}{3}}^{π} sin (u) \frac{3}{π} d u

Take the constant out:

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

= 3 \cdot \frac{3}{π} \cdot \int_{\frac{2 π}{3}}^{π} sin (u) d u

Use the common integral:

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

= 3 \cdot \frac{3}{π} [- cos (u)]_{\frac{2 π}{3}}^{π}

Simplify

3 \cdot \frac{3}{π} [- cos (u)]_{\frac{2 π}{3}}^{π} : \frac{9}{π} [- cos (u)]_{\frac{2 π}{3}}^{π}

= \frac{9}{π} [- cos (u)]_{\frac{2 π}{3}}^{π}

Compute the boundaries:

\frac{1}{2}

= \frac{9}{π} \cdot \frac{1}{2}

Simplify

= \frac{9}{2 π}

\int_{0}^{π} 3 x d x

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What is the integral from 2 to 3 of 3sin(pi/3 x) ?
The integral from 2 to 3 of 3sin(pi/3 x) is 9/(2pi)