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

The integral from 0 to 1 of (x^3-1)e^x is -3e+7

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

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

- 3 e + 7

Decimal

- 1.15484 \dots

Solution steps

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

Expand

(x^{3} - 1) e^{x} : e^{x} x^{3} - e^{x}

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

\int_{0}^{1} e^{x} x^{3} d x = - 2 e + 6

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

= - 2 e + 6 - (e - 1)

Simplify

- 2 e + 6 - (e - 1) : - 3 e + 7

= - 3 e + 7

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What is the integral from 0 to 1 of (x^3-1)e^x ?
The integral from 0 to 1 of (x^3-1)e^x is -3e+7