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

The integral from 0 to 2 of te^{t^2} is (e^4-1)/2

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

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

\frac{e ^{4} - 1}{2}

Decimal Notation

26.79907 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{4} \frac{e ^{u}}{2} d u

= \int_{0}^{4} \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}^{4} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= \frac{1}{2} (e^{4} - 1)

Simplify

= \frac{e ^{4} - 1}{2}

\int \frac{3 x}{( 2 + 3 x ^{2} ) ^{3}} d x

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

\frac{\partial}{\partial x} (x^{yy - xx})

x \to 1 lim (\frac{2 x}{x ^{3} + 1})

\frac{d}{d x} ((\frac{- 2 x}{1 + x ^{2}})^{2})

What is the integral from 0 to 2 of te^{t^2} ?
The integral from 0 to 2 of te^{t^2} is (e^4-1)/2