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

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

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integral from 0 to 2 of te^t

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

e^{2} + 1

Decimal Notation

8.38905 \dots

Solution steps

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

Apply Integration By Parts

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

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

\int e^{t} d t = e^{t}

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

Compute the boundaries:

e^{2} + 1

= e^{2} + 1

sin^{2} (x), sin^{3} (x), 0, π

\int 3 sin (x) + 2 - 3 e^{x} d x

f (x) = 4 x (1 - x), g (x) = 4 x - 4

\frac{d}{d x} (\frac{6 x ^{3} - 5 x ^{2} - 2 x}{4 x - 3 x ^{2}})

\frac{d}{d x} (5 x + \frac{100}{x})

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