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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{t^{2}} - 1

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

\int_{0}^{3} t e^{- s t} d t

\int_{0}^{9} (\frac{10}{( 1 + x ) ^{2}}) d x

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

\int_{4}^{9} 2 t (10 t + 3) d t

\int_{0}^{2} 3 x - x^{2} d x

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