What is the integral from 0 to 2 of 2xe^x-e^x+e^{-x} ?

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

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integral from 0 to 2 of 2xe^x-e^x+e^{-x}

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

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

Decimal

11.25372 \dots

Solution steps

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

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

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

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

= 2 (e^{2} + 1) - (e^{2} - 1) + 1 - \frac{1}{e ^{2}}

Simplify

2 (e^{2} + 1) - (e^{2} - 1) + 1 - \frac{1}{e ^{2}} : 4 + e^{2} - \frac{1}{e ^{2}}

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

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

\int_{- 4}^{4} - e^{(\frac{1}{4}) x} d x

\int_{4}^{8} 10 - x d x

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

\int_{- 2}^{0} 4 x - x^{2} - 6 x d x

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