What is the integral from 1 to 3 of (4x-2e^{3x}) ?

The integral from 1 to 3 of (4x-2e^{3x}) is 16-(2(e^9-e^3))/3

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integral from 1 to 3 of (4x-2e^{3x})

\int_{1}^{3} (4 x - 2 e^{3 x}) d x

16 - \frac{2 ( e ^{9} - e ^{3} )}{3}

Decimal

- 5372.66559 \dots

Solution steps

\int_{1}^{3} 4 x - 2 e^{3 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

= \int_{1}^{3} 4 x d x - \int_{1}^{3} 2 e^{3 x} d x

\int_{1}^{3} 4 x d x = 16

\int_{1}^{3} 2 e^{3 x} d x = \frac{2 ( e ^{9} - e ^{3} )}{3}

= 16 - \frac{2 ( e ^{9} - e ^{3} )}{3}

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What is the integral from 1 to 3 of (4x-2e^{3x}) ?
The integral from 1 to 3 of (4x-2e^{3x}) is 16-(2(e^9-e^3))/3