What is the integral from 0 to 1 of x^2e^{-x^3} ?

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

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integral from 0 to 1 of x^2e^{-x^3}

\int_{0}^{1} x^{2} e^{- x^{3}} d x

\frac{e - 1}{3 e}

Decimal

0.21070 \dots

Solution steps

\int_{0}^{1} x^{2} e^{- x^{3}} d x

Apply u-substitution

: \int_{0}^{- 1} - \frac{e ^{u}}{3} d u

= \int_{0}^{- 1} - \frac{e ^{u}}{3} d u

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

= - \int_{- 1}^{0} - \frac{e ^{u}}{3} d u

Take the constant out:

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

= - (- \frac{1}{3} \cdot \int_{- 1}^{0} e^{u} d u)

Use the common integral:

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

= - (- \frac{1}{3} [e^{u}]_{- 1}^{0})

Simplify

= \frac{1}{3} [e^{u}]_{- 1}^{0}

Compute the boundaries:

1 - \frac{1}{e}

= \frac{1}{3} (1 - \frac{1}{e})

Simplify

= \frac{e - 1}{3 e}

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{3}} d x

\int_{0}^{3} x e^{2 x} d x

\int_{0}^{4} \frac{1}{( x - 3 ) ^{2}} d x

\int_{0}^{2} - x^{2} + 2 x d x

\int_{0}^{0.3} sin (x^{2}) d x

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