What is the integral from 0 to 1 of t^2e^t ?

The integral from 0 to 1 of t^2e^t is e-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of t^2e^t

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

e - 2

Decimal Notation

0.71828 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int 2 t e^{t} d t = 2 (e^{t} t - e^{t})

= [t^{2} e^{t} - 2 (e^{t} t - e^{t})]_{0}^{1}

Compute the boundaries:

e - 2

= e - 2

t y^{^{'}} - y = t^{2} e^{- t}

\int \frac{- 1 + x}{x ^{2}} d x

\frac{\partial}{\partial x} (3 x^{2} y^{4} - 4)

\int \frac{1}{sin ( x ) + cos ( x )} d x

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

What is the integral from 0 to 1 of t^2e^t ?
The integral from 0 to 1 of t^2e^t is e-2