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

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

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

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

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

Decimal Notation

2.33538 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

e^{2} - e

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

Simplify

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

\int - 2 tan (2 y) d y

\frac{\partial}{\partial x} (9 - x^{2} + x y - 3 y^{2})

12 - x^{2}, x^{2} - 6, [- 3, 3]

y = \frac{1}{x}, (2, \frac{1}{2})

\frac{\partial}{\partial x} (e^{s i n (x)})

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