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

The integral from 0 to x of te^{t^2} is 1/2 (e^{x^2}-1)

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integral from 0 to x of te^{t^2}

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

\frac{1}{2} (e^{x^{2}} - 1)

Solution steps

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

Apply u-substitution

: \int_{0}^{x^{2}} \frac{e ^{u}}{2} d u

= \int_{0}^{x^{2}} \frac{e ^{u}}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{x^{2}} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{0}^{x^{2}}

Compute the boundaries:

e^{x^{2}} - 1

= \frac{1}{2} (e^{x^{2}} - 1)

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