What is the integral from 0 to 11 of t^2e^{2t} ?

The integral from 0 to 11 of t^2e^{2t} is (221e^{22}-1)/4

Popular >

integral from 0 to 11 of t^2e^{2t}

\int_{0}^{11} t^{2} e^{2 t} d t

\frac{221 e ^{22} - 1}{4}

Decimal Notation

198066434748.5

Solution steps

\int_{0}^{11} t^{2} e^{2 t} d t

Apply Integration By Parts

: [\frac{1}{2} e^{2 t} t^{2} - \int e^{2 t} t d t]_{0}^{11}

= [\frac{1}{2} e^{2 t} t^{2} - \int e^{2 t} t d t]_{0}^{11}

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

= [\frac{1}{2} e^{2 t} t^{2} - \frac{1}{4} (e^{2 t} \cdot 2 t - e^{2 t})]_{0}^{11}

Compute the boundaries:

\frac{221 e ^{22}}{4} - \frac{1}{4}

= \frac{221 e ^{22}}{4} - \frac{1}{4}

Simplify

= \frac{221 e ^{22} - 1}{4}

\int_{0}^{1} e^{x^{2}} 2 x d x

\int_{1}^{3} x^{3} (ln (x))^{2} d x

\int_{- 2}^{4} x^{2} sin (x^{3} + 3) d x

\int_{0}^{3} 5 e^{x} d x

\int_{1}^{3} \frac{2 x}{1 + x ^{2}} d x

What is the integral from 0 to 11 of t^2e^{2t} ?
The integral from 0 to 11 of t^2e^{2t} is (221e^{22}-1)/4