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

The integral from 0 to 2 of x^2e^{2x} is 2e^4+(-3e^4-1)/4

Popular >

integral from 0 to 2 of x^2e^{2x}

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

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

Decimal Notation

67.99768 \dots

Solution steps

\int_{0}^{2} 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]_{0}^{2}

= [\frac{1}{2} e^{2 x} x^{2} - \int e^{2 x} x d x]_{0}^{2}

\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})]_{0}^{2}

Compute the boundaries:

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

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

Simplify

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

\int_{0}^{2} (6 x^{2} + 4 x + 2) d x

\int_{- 3}^{3} (x^{2} - 5 x + 1) d x

\int_{0}^{1} e^{y^{3}} y^{2} d y

\int_{2}^{2} cos (e^{t} + ln (t)) d t

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

What is the integral from 0 to 2 of x^2e^{2x} ?
The integral from 0 to 2 of x^2e^{2x} is 2e^4+(-3e^4-1)/4