What is the integral from 0 to 1 of 3te^{8t^2} ?

The integral from 0 to 1 of 3te^{8t^2} is (3(e^8-1))/(16)

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integral from 0 to 1 of 3te^{8t^2}

\int_{0}^{1} 3 t e^{8 t^{2}} d t

\frac{3 ( e ^{8} - 1 )}{16}

Decimal

558.74212 \dots

Solution steps

\int_{0}^{1} 3 t e^{8 t^{2}} d t

Take the constant out:

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

= 3 \cdot \int_{0}^{1} t e^{8 t^{2}} d t

Apply u-substitution

: \int_{0}^{8} \frac{e ^{u}}{16} d u

= 3 \cdot \int_{0}^{8} \frac{e ^{u}}{16} d u

Take the constant out:

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

= 3 \cdot \frac{1}{16} \cdot \int_{0}^{8} e^{u} d u

Use the common integral:

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

= 3 \cdot \frac{1}{16} [e^{u}]_{0}^{8}

Simplify

3 \cdot \frac{1}{16} [e^{u}]_{0}^{8} : \frac{3}{16} [e^{u}]_{0}^{8}

= \frac{3}{16} [e^{u}]_{0}^{8}

Compute the boundaries:

e^{8} - 1

= \frac{3}{16} (e^{8} - 1)

Simplify

= \frac{3 ( e ^{8} - 1 )}{16}

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What is the integral from 0 to 1 of 3te^{8t^2} ?
The integral from 0 to 1 of 3te^{8t^2} is (3(e^8-1))/(16)