What is the integral from 0 to 2 of 2xe^{x^2} ?

The integral from 0 to 2 of 2xe^{x^2} is e^4-1

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integral from 0 to 2 of 2xe^{x^2}

\int_{0}^{2} 2 x e^{x^{2}} d x

e^{4} - 1

Decimal Notation

53.59815 \dots

Solution steps

\int_{0}^{2} 2 x e^{x^{2}} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} x e^{x^{2}} d x

Apply u-substitution

: \int_{0}^{4} \frac{e ^{u}}{2} d u

= 2 \cdot \int_{0}^{4} \frac{e ^{u}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{2} [e^{u}]_{0}^{4}

Simplify

2 \cdot \frac{1}{2} [e^{u}]_{0}^{4} : [e^{u}]_{0}^{4}

= [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= e^{4} - 1

\int \frac{3}{x} + 3 x d x

\frac{\partial}{\partial t} (t e^{t})

x = 4 y, x = y^{2} - 12

3 x^{- 4}

What is the integral from 0 to 2 of 2xe^{x^2} ?
The integral from 0 to 2 of 2xe^{x^2} is e^4-1