What is the integral from-3 to 0 of (6x-e^x) ?

The integral from-3 to 0 of (6x-e^x) is 1/(e^3)-28

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integral from-3 to 0 of (6x-e^x)

\int_{- 3}^{0} (6 x - e^{x}) d x

\frac{1}{e ^{3}} - 28

Decimal

- 27.95021 \dots

Solution steps

\int_{- 3}^{0} 6 x - e^{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_{- 3}^{0} 6 x d x - \int_{- 3}^{0} e^{x} d x

\int_{- 3}^{0} 6 x d x = - 27

\int_{- 3}^{0} e^{x} d x = 1 - \frac{1}{e ^{3}}

= - 27 - (1 - \frac{1}{e ^{3}})

Simplify

= \frac{1}{e ^{3}} - 28

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What is the integral from-3 to 0 of (6x-e^x) ?
The integral from-3 to 0 of (6x-e^x) is 1/(e^3)-28