What is the integral from 0 to 1 of te^{6t^2} ?

The integral from 0 to 1 of te^{6t^2} is (e^6-1)/(12)

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integral from 0 to 1 of te^{6t^2}

\int_{0}^{1} t e^{6 t^{2}} d t

\frac{e ^{6} - 1}{12}

Decimal

33.53573 \dots

Solution steps

\int_{0}^{1} t e^{6 t^{2}} d t

Apply u-substitution

: \int_{0}^{6} \frac{e ^{u}}{12} d u

= \int_{0}^{6} \frac{e ^{u}}{12} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int_{0}^{6} e^{u} d u

Use the common integral:

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

= \frac{1}{12} [e^{u}]_{0}^{6}

Compute the boundaries:

e^{6} - 1

= \frac{1}{12} (e^{6} - 1)

Simplify

= \frac{e ^{6} - 1}{12}

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What is the integral from 0 to 1 of te^{6t^2} ?
The integral from 0 to 1 of te^{6t^2} is (e^6-1)/(12)