What is the integral from 0 to 1 of 2e^{r^2}r ?

The integral from 0 to 1 of 2e^{r^2}r is e-1

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integral from 0 to 1 of 2e^{r^2}r

\int_{0}^{1} 2 e^{r^{2}} r d r

e - 1

Decimal

1.71828 \dots

Solution steps

\int_{0}^{1} 2 e^{r^{2}} r d r

Take the constant out:

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

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

Apply u-substitution

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

= 2 \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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

e - 1

= e - 1

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