What is the integral from 2 to 3 of (x-3)e^{2x} ?

The integral from 2 to 3 of (x-3)e^{2x} is (-e^6+3e^4)/4

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integral from 2 to 3 of (x-3)e^{2x}

\int_{2}^{3} (x - 3) e^{2 x} d x

\frac{- e ^{6} + 3 e ^{4}}{4}

Decimal Notation

- 59.90858 \dots

Solution steps

\int_{2}^{3} (x - 3) e^{2 x} d x

Apply Integration By Parts

: [\frac{1}{2} e^{2 x} (x - 3) - \int \frac{1}{2} e^{2 x} d x]_{2}^{3}

= [\frac{1}{2} e^{2 x} (x - 3) - \int \frac{1}{2} e^{2 x} d x]_{2}^{3}

\int \frac{1}{2} e^{2 x} d x = \frac{1}{4} e^{2 x}

= [\frac{1}{2} e^{2 x} (x - 3) - \frac{1}{4} e^{2 x}]_{2}^{3}

Simplify

[\frac{1}{2} e^{2 x} (x - 3) - \frac{1}{4} e^{2 x}]_{2}^{3} : [\frac{1}{4} (2 e^{2 x} x - 7 e^{2 x})]_{2}^{3}

= [\frac{1}{4} (2 e^{2 x} x - 7 e^{2 x})]_{2}^{3}

Compute the boundaries:

\frac{- e ^{6} + 3 e ^{4}}{4}

= \frac{- e ^{6} + 3 e ^{4}}{4}

\int_{1}^{4} \frac{1}{2 x} d x

\int_{- 1}^{1} (∣ x ∣ - 1) d x

\int_{3}^{4} x^{2} - 2 x - 3 d x

\int_{0}^{\frac{π}{2}} \frac{1}{1 + cos ( x )} d x

\int_{- 1}^{2} 3 x d x

What is the integral from 2 to 3 of (x-3)e^{2x} ?
The integral from 2 to 3 of (x-3)e^{2x} is (-e^6+3e^4)/4