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

The integral from 1 to 3 of xe^{x^2} is (e^9-e)/2

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

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

\frac{e ^{9} - e}{2}

Decimal

4050.18282 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{9} \frac{e ^{u}}{2} d u

= \int_{1}^{9} \frac{e ^{u}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{9} e^{u} d u

Use the common integral:

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

= \frac{1}{2} [e^{u}]_{1}^{9}

Compute the boundaries:

e^{9} - e

= \frac{1}{2} (e^{9} - e)

Simplify

= \frac{e ^{9} - e}{2}

ln (\frac{3 x ^{4} + 2}{4 x})

\frac{\partial}{\partial x} (y^{3} e^{2 x} + 5 sin (x y))

\frac{d}{d x} (x^{2} + 3 x - x)

y^{^{''}} + 6 y = 0, y^{^{'}} (1) = 5

\frac{d y}{d x} = \frac{( x ^{2} + 1 )}{y ^{2} + 1}, y (0) = 1

What is the integral from 1 to 3 of xe^{x^2} ?
The integral from 1 to 3 of xe^{x^2} is (e^9-e)/2