What is the integral from 0 to 4 of (3sqrt(t)-2e^t) ?

The integral from 0 to 4 of (3sqrt(t)-2e^t) is -2e^4+18

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integral from 0 to 4 of (3sqrt(t)-2e^t)

\int_{0}^{4} (3 t - 2 e^{t}) d t

- 2 e^{4} + 18

Decimal

- 91.19630 \dots

Solution steps

\int_{0}^{4} 3 t - 2 e^{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}^{4} 3 t d t - \int_{0}^{4} 2 e^{t} d t

\int_{0}^{4} 3 t d t = 16

\int_{0}^{4} 2 e^{t} d t = 2 (e^{4} - 1)

= 16 - 2 (e^{4} - 1)

Simplify

16 - 2 (e^{4} - 1) : - 2 e^{4} + 18

= - 2 e^{4} + 18

\frac{d}{d x} (x^{2} - 16)

4 e^{- t}

\int_{- 2}^{5} (x^{2} - 4) d x

\frac{d}{d x} ((5 x^{4} - 6 x^{3} + x)^{5})

\int_{0}^{2} \frac{4}{x ^{2}} d x

What is the integral from 0 to 4 of (3sqrt(t)-2e^t) ?
The integral from 0 to 4 of (3sqrt(t)-2e^t) is -2e^4+18