What is the integral from 0 to 2 of 6xe^{x^2} ?

The integral from 0 to 2 of 6xe^{x^2} is 3(e^4-1)

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integral from 0 to 2 of 6xe^{x^2}

\int_{0}^{2} 6 x e^{x^{2}} d x

3 (e^{4} - 1)

Decimal

160.79445 \dots

Solution steps

\int_{0}^{2} 6 x e^{x^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{2} x e^{x^{2}} d x

Apply u-substitution

= 6 \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

= 6 \cdot \frac{1}{2} \cdot \int_{0}^{4} e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} [e^{u}]_{0}^{4}

Simplify

6 \cdot \frac{1}{2} [e^{u}]_{0}^{4} : 3 [e^{u}]_{0}^{4}

= 3 [e^{u}]_{0}^{4}

Compute the boundaries:

e^{4} - 1

= 3 (e^{4} - 1)

What is the integral from 0 to 2 of 6xe^{x^2} ?
The integral from 0 to 2 of 6xe^{x^2} is 3(e^4-1)