What is the integral from 1 to sqrt(2 of)xe^{4x^2} ?

The integral from 1 to sqrt(2 of)xe^{4x^2} is (e^8-e^4)/8

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integral from 1 to sqrt(2 of)xe^{4x^2}

\int_{1}^{2} x e^{4 x^{2}} d x

\frac{e ^{8} - e ^{4}}{8}

Decimal

365.79497 \dots

Solution steps

\int_{1}^{2} x e^{4 x^{2}} d x

Apply u-substitution

: \int_{4}^{8} \frac{e ^{u}}{8} d u

= \int_{4}^{8} \frac{e ^{u}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int_{4}^{8} e^{u} d u

Use the common integral:

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

= \frac{1}{8} [e^{u}]_{4}^{8}

Compute the boundaries:

e^{8} - e^{4}

= \frac{1}{8} (e^{8} - e^{4})

Simplify

= \frac{e ^{8} - e ^{4}}{8}

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What is the integral from 1 to sqrt(2 of)xe^{4x^2} ?
The integral from 1 to sqrt(2 of)xe^{4x^2} is (e^8-e^4)/8