What is the integral from 1 to sqrt(2 of)xe^{3x^2} ?

The integral from 1 to sqrt(2 of)xe^{3x^2} is (e^6-e^3)/6

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integral from 1 to sqrt(2 of)xe^{3x^2}

\int_{1}^{2} x e^{3 x^{2}} d x

\frac{e ^{6} - e ^{3}}{6}

Decimal Notation

63.89054 \dots

Solution steps

\int_{1}^{2} x e^{3 x^{2}} d x

Apply u-substitution

: \int_{3}^{6} \frac{e ^{u}}{6} d u

= \int_{3}^{6} \frac{e ^{u}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int_{3}^{6} e^{u} d u

Use the common integral:

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

= \frac{1}{6} [e^{u}]_{3}^{6}

Compute the boundaries:

e^{6} - e^{3}

= \frac{1}{6} (e^{6} - e^{3})

Simplify

= \frac{e ^{6} - e ^{3}}{6}

\frac{d y}{d x} = \frac{y}{x}

\frac{\partial}{\partial y} (\frac{cy}{x ^{2} + y ^{2}})

(4 + x) y^{^{'}} = y^{3}

x \to 2 lim (\frac{1}{x - 1})

i = 1 \sum \infty (i) \cdot \frac{1}{6 ^{i}}

What is the integral from 1 to sqrt(2 of)xe^{3x^2} ?
The integral from 1 to sqrt(2 of)xe^{3x^2} is (e^6-e^3)/6