What is the integral from 1 to sqrt(2 of)xe^{8x^2} ?

The integral from 1 to sqrt(2 of)xe^{8x^2} is (e^{16}-e^8)/(16)

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integral from 1 to sqrt(2 of)xe^{8x^2}

\int_{1}^{2} x e^{8 x^{2}} d x

\frac{e ^{16} - e ^{8}}{16}

Decimal Notation

555195.59765 \dots

Solution steps

\int_{1}^{2} x e^{8 x^{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{16} - e^{8}

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

Simplify

= \frac{e ^{16} - e ^{8}}{16}

x y^{^{'}} + 2 y = - e^{x}

y^{^{'}} = - 0.5 y

f (x) = x^{2} (- 3 x^{2} - 2)

\int \frac{1}{x ( x + 3 )} d x

8 e^{- x}

What is the integral from 1 to sqrt(2 of)xe^{8x^2} ?
The integral from 1 to sqrt(2 of)xe^{8x^2} is (e^{16}-e^8)/(16)