What is the integral from 0 to 4 of (4e^{-0.25x}-6x) ?

The integral from 0 to 4 of (4e^{-0.25x}-6x) is -(16)/e-32

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integral from 0 to 4 of (4e^{-0.25x}-6x)

\int_{0}^{4} (4 e^{- 0.25 x} - 6 x) d x

- \frac{16}{e} - 32

Decimal

- 37.88607 \dots

Solution steps

\int_{0}^{4} 4 e^{- 0.25 x} - 6 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}^{4} 4 e^{- 0.25 x} d x - \int_{0}^{4} 6 x d x

\int_{0}^{4} 4 e^{- 0.25 x} d x = 16 (1 - \frac{1}{e})

\int_{0}^{4} 6 x d x = 48

= 16 (1 - \frac{1}{e}) - 48

Simplify

16 (1 - \frac{1}{e}) - 48 : - \frac{16}{e} - 32

= - \frac{16}{e} - 32

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\frac{\partial}{\partial x} (9 sin (x y))

What is the integral from 0 to 4 of (4e^{-0.25x}-6x) ?
The integral from 0 to 4 of (4e^{-0.25x}-6x) is -(16)/e-32