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

The integral from-2 to-1 of (2x-1)e^{2x} is -2/(e^2)+3/(e^4)

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

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

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

Decimal Notation

- 0.21572 \dots

Solution steps

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

Apply Integration By Parts

: [e^{2 x} x - \frac{1}{2} e^{2 x} - \int e^{2 x} d x]_{- 2}^{- 1}

= [e^{2 x} x - \frac{1}{2} e^{2 x} - \int e^{2 x} d x]_{- 2}^{- 1}

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

= [e^{2 x} x - \frac{1}{2} e^{2 x} - \frac{1}{2} e^{2 x}]_{- 2}^{- 1}

Add similar elements:

- \frac{1}{2} e^{2 x} - \frac{1}{2} e^{2 x} = - e^{2 x}

= [e^{2 x} x - e^{2 x}]_{- 2}^{- 1}

Compute the boundaries:

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

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

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What is the integral from-2 to-1 of (2x-1)e^{2x} ?
The integral from-2 to-1 of (2x-1)e^{2x} is -2/(e^2)+3/(e^4)