What is the integral from 2 to 4 of x^2e^{2x} ?

The integral from 2 to 4 of x^2e^{2x} is 8e^8-2e^4+(-7e^8+3e^4)/4

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integral from 2 to 4 of x^2e^{2x}

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

8 e^{8} - 2 e^{4} + \frac{- 7 e ^{8} + 3 e ^{4}}{4}

Decimal Notation

18562.73973 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{1}{2} e^{2 x} x^{2} - \int e^{2 x} x d x]_{2}^{4}

= [\frac{1}{2} e^{2 x} x^{2} - \int e^{2 x} x d x]_{2}^{4}

\int e^{2 x} x d x = \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

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

Compute the boundaries:

8 e^{8} - 2 e^{4} + \frac{- 7 e ^{8} + 3 e ^{4}}{4}

= 8 e^{8} - 2 e^{4} + \frac{- 7 e ^{8} + 3 e ^{4}}{4}

(e^{e x} - x^{2} y) = 1

\int (\frac{x ^{2} - 1}{x ^{4}}) d x

(x^{2} + 5) y^{^{'}} = x y

y = 3 x^{2}, (0, 1)

\frac{d}{d x} ((x^{3} + 9) e^{x})

What is the integral from 2 to 4 of x^2e^{2x} ?
The integral from 2 to 4 of x^2e^{2x} is 8e^8-2e^4+(-7e^8+3e^4)/4