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

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

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integral from 0 to 1 of (x+e^x)

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

e - \frac{1}{2}

Decimal

2.21828 \dots

Solution steps

\int_{0}^{1} 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_{0}^{1} x d x + \int_{0}^{1} e^{x} d x

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

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

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

Combine the fractions

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

= e - \frac{1}{2}

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