What is the integral from 0 to 5 of (2e^{0.2x}-x) ?

The integral from 0 to 5 of (2e^{0.2x}-x) is 10e-45/2

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integral from 0 to 5 of (2e^{0.2x}-x)

\int_{0}^{5} (2 e^{0.2 x} - x) d x

10 e - \frac{45}{2}

Decimal Notation

4.68281 \dots

Solution steps

\int_{0}^{5} 2 e^{0.2 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}^{5} 2 e^{0.2 x} d x - \int_{0}^{5} x d x

\int_{0}^{5} 2 e^{0.2 x} d x = 10 (e - 1)

\int_{0}^{5} x d x = \frac{25}{2}

= 10 (e - 1) - \frac{25}{2}

Simplify

10 (e - 1) - \frac{25}{2} : 10 e - \frac{45}{2}

= 10 e - \frac{45}{2}

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What is the integral from 0 to 5 of (2e^{0.2x}-x) ?
The integral from 0 to 5 of (2e^{0.2x}-x) is 10e-45/2