What is the integral from 0 to 1 of [e^x-x] ?

The integral from 0 to 1 of [e^x-x] is e-3/2

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integral from 0 to 1 of [e^x-x]

\int_{0}^{1} [e^{x} - x] d x

e - \frac{3}{2}

Decimal

1.21828 \dots

Solution steps

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

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

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

= e - 1 - \frac{1}{2}

Combine the fractions

- 1 - \frac{1}{2} : - \frac{3}{2}

= e - \frac{3}{2}

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What is the integral from 0 to 1 of [e^x-x] ?
The integral from 0 to 1 of [e^x-x] is e-3/2