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

The integral from 0 to 1/2 of 2xe^{-x^2} is 1-1/(\sqrt[4]{e)}

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

\int_{0}^{\frac{1}{2}} 2 x e^{- x^{2}} d x

Solution steps

\int_{0}^{\frac{1}{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}^{\frac{1}{2}} x e^{- x^{2}} d x

Apply u-substitution

: \int_{0}^{- \frac{1}{4}} - \frac{e ^{u}}{2} d u

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Use the common integral:

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

= 2 (- (- \frac{1}{2} [e^{u}]_{- \frac{1}{4}}^{0}))

Simplify

2 (- (- \frac{1}{2} [e^{u}]_{- \frac{1}{4}}^{0})) : [e^{u}]_{- \frac{1}{4}}^{0}

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

Compute the boundaries:

\int_{0}^{3} 4 x e^{- x} d x

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\int_{\frac{1}{2}}^{\frac{e}{2}} x ln (2 x) d x

\int_{0}^{π} cos^{3} (5 x) d x

\int_{0}^{1} x \cdot 3 (1 - x)^{2} d x

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