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

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

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integral from 0 to 1 of 4xe^{x^2}

\int_{0}^{1} 4 x e^{x^{2}} d x

2 (e - 1)

Decimal

3.43656 \dots

Solution steps

\int_{0}^{1} 4 x e^{x^{2}} d x

Take the constant out:

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

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

Apply u-substitution

: \int_{0}^{1} \frac{e ^{u}}{2} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Simplify

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

= 2 [e^{u}]_{0}^{1}

Compute the boundaries:

e - 1

= 2 (e - 1)

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