What is the integral from 0 to 4 of (e^{-2x}+1) ?

The integral from 0 to 4 of (e^{-2x}+1) is (e^8-1)/(2e^8)+4

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integral from 0 to 4 of (e^{-2x}+1)

\int_{0}^{4} (e^{- 2 x} + 1) d x

\frac{e ^{8} - 1}{2 e ^{8}} + 4

Decimal Notation

4.49983 \dots

Solution steps

\int_{0}^{4} e^{- 2 x} + 1 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} e^{- 2 x} d x + \int_{0}^{4} 1 d x

\int_{0}^{4} e^{- 2 x} d x = \frac{e ^{8} - 1}{2 e ^{8}}

\int_{0}^{4} 1 d x = 4

= \frac{e ^{8} - 1}{2 e ^{8}} + 4

\frac{d}{d x} (y^{x})

f (x) = 7 x + 6

\int \frac{1}{2 tan ( x )} d x

\frac{s ^{2} + 3 s + 5}{s ^{3} ( s + 2 )}

\frac{d y}{d x} = x

What is the integral from 0 to 4 of (e^{-2x}+1) ?
The integral from 0 to 4 of (e^{-2x}+1) is (e^8-1)/(2e^8)+4