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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{2 x^{2}} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

e^{2 x^{2}} - 1

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

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

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\int_{0}^{\frac{π}{3}} 5 sin (\frac{x}{2}) d x

\int_{0}^{120} 5 - 5 e^{- 0.2 x} d x

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