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

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

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

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

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

Decimal

53.59815 \dots

Solution steps

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

Expand

(x^{2} - x + 1) e^{2 x} : e^{2 x} x^{2} - e^{2 x} x + e^{2 x}

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

\int_{0}^{2} e^{2 x} x^{2} d x = 2 e^{4} + \frac{- 3 e ^{4} - 1}{4}

\int_{0}^{2} e^{2 x} x d x = \frac{3 e ^{4} + 1}{4}

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

= 2 e^{4} + \frac{- 3 e ^{4} - 1}{4} - \frac{3 e ^{4} + 1}{4} + \frac{e ^{4} - 1}{2}

Simplify

2 e^{4} + \frac{- 3 e ^{4} - 1}{4} - \frac{3 e ^{4} + 1}{4} + \frac{e ^{4} - 1}{2} : 2 e^{4} + \frac{- 4 e ^{4} - 4}{4}

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

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