What is the integral from 0 to 2 of 5(1-e^x) ?

The integral from 0 to 2 of 5(1-e^x) is 5(-e^2+3)

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integral from 0 to 2 of 5(1-e^x)

\int_{0}^{2} 5 (1 - e^{x}) d x

5 (- e^{2} + 3)

Decimal

- 21.94528 \dots

Solution steps

\int_{0}^{2} 5 (1 - e^{x}) d x

Take the constant out:

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

= 5 \cdot \int_{0}^{2} 1 - 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

= 5 (\int_{0}^{2} 1 d x - \int_{0}^{2} e^{x} d x)

\int_{0}^{2} 1 d x = 2

\int_{0}^{2} e^{x} d x = e^{2} - 1

= 5 (2 - (e^{2} - 1))

Simplify

= 5 (- e^{2} + 3)

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What is the integral from 0 to 2 of 5(1-e^x) ?
The integral from 0 to 2 of 5(1-e^x) is 5(-e^2+3)