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

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

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

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

\frac{e - 1}{3 e}

Decimal

0.21070 \dots

Solution steps

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

Apply u-substitution

= \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}

Compute the boundaries:

1 - \frac{1}{e}

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

Simplify

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

\frac{\partial}{\partial x} (x^{2} + y^{3} + w^{4})

a re a y = ∣ 3 x ∣, y = x^{2} - 4

\int \frac{1}{1 - 4 x} d x

x \to - 1 lim (\frac{x ^{2} + 5}{x + 6})

d er i v a t i v e (2 x + 3)^{1.4}

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