What is the integral from 0 to pi of (9-sin(4x))/4 ?

The integral from 0 to pi of (9-sin(4x))/4 is (9pi)/4

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integral from 0 to pi of (9-sin(4x))/4

\int_{0}^{π} \frac{9 - sin ( 4 x )}{4} d x

\frac{9 π}{4}

Decimal

7.06858 \dots

Solution steps

\int_{0}^{π} \frac{9 - sin ( 4 x )}{4} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{π} 9 - sin (4 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int_{0}^{π} 9 d x - \int_{0}^{π} sin (4 x) d x)

\int_{0}^{π} 9 d x = 9 π

\int_{0}^{π} sin (4 x) d x = 0

= \frac{1}{4} (9 π - 0)

Simplify

\frac{1}{4} (9 π - 0) : \frac{9 π}{4}

= \frac{9 π}{4}

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What is the integral from 0 to pi of (9-sin(4x))/4 ?
The integral from 0 to pi of (9-sin(4x))/4 is (9pi)/4