What is the integral from-1 to 0 of (8x-e^x) ?

The integral from-1 to 0 of (8x-e^x) is 1/e-5

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integral from-1 to 0 of (8x-e^x)

\int_{- 1}^{0} (8 x - e^{x}) d x

\frac{1}{e} - 5

Decimal Notation

- 4.63212 \dots

Solution steps

\int_{- 1}^{0} 8 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_{- 1}^{0} 8 x d x - \int_{- 1}^{0} e^{x} d x

\int_{- 1}^{0} 8 x d x = - 4

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

= - 4 - (1 - \frac{1}{e})

Simplify

= \frac{1}{e} - 5

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What is the integral from-1 to 0 of (8x-e^x) ?
The integral from-1 to 0 of (8x-e^x) is 1/e-5