What is the integral from 0 to 3 of 5(1-e^y) ?

The integral from 0 to 3 of 5(1-e^y) is 5(-e^3+4)

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integral from 0 to 3 of 5(1-e^y)

\int_{0}^{3} 5 (1 - e^{y}) d y

5 (- e^{3} + 4)

Solution steps

\int_{0}^{3} 5 (1 - e^{y}) d y

Take the constant out:

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

= 5 \cdot \int_{0}^{3} 1 - e^{y} d y

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}^{3} 1 d y - \int_{0}^{3} e^{y} d y)

\int_{0}^{3} 1 d y = 3

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

= 5 (3 - (e^{3} - 1))

Simplify

= 5 (- e^{3} + 4)

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\int_{0}^{2} 2 x^{2} - 4 x d x

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What is the integral from 0 to 3 of 5(1-e^y) ?
The integral from 0 to 3 of 5(1-e^y) is 5(-e^3+4)