What is the integral from-pi to 1 of (3x+pi) ?

The integral from-pi to 1 of (3x+pi) is (3(1-pi^2))/2+pi+pi^2

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integral from-pi to 1 of (3x+pi)

\int_{- π}^{1} (3 x + π) d x

\frac{3 ( 1 - π ^{2} )}{2} + π + π^{2}

Decimal

- 0.29320 \dots

Solution steps

\int_{- π}^{1} 3 x + π 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

= \int_{- π}^{1} 3 x d x + \int_{- π}^{1} π d x

\int_{- π}^{1} 3 x d x = \frac{3 ( 1 - π ^{2} )}{2}

\int_{- π}^{1} π d x = π + π^{2}

= \frac{3 ( 1 - π ^{2} )}{2} + π + π^{2}

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What is the integral from-pi to 1 of (3x+pi) ?
The integral from-pi to 1 of (3x+pi) is (3(1-pi^2))/2+pi+pi^2