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

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

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integral from 0 to 5 of 2e^{0.2t}-t

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

10 e - \frac{45}{2}

Decimal

4.68281 \dots

Solution steps

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

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 t} d t - \int_{0}^{5} t d t

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

\int_{0}^{5} t d t = \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^{0.2t}-t ?
The integral from 0 to 5 of 2e^{0.2t}-t is 10e-45/2