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

The integral from 0 to x of te^t is xe^x-e^x+1

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integral from 0 to x of te^t

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

x e^{x} - e^{x} + 1

Solution steps

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

x e^{x} - e^{x} + 1

= x e^{x} - e^{x} + 1

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

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\int_{\frac{1}{2}}^{1} y cos^{2} (x y) d x

\int_{- 2}^{2} (4 - y^{2}) d y

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