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

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

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

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

\frac{e - 1}{3}

Solution steps

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

Apply u-substitution

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

= \int_{0}^{1} \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_{0}^{1} e^{u} d u

Use the common integral:

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

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

Compute the boundaries:

e - 1

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

Simplify

= \frac{e - 1}{3}

y^{5} \cdot (y^{^{'}} - y) = e^{6 x}

\frac{d y}{d x} = - \frac{2 y ^{2}}{x ^{2}}

(x + 4 x) (x^{\frac{3}{2}} - x)

y = ln ∣ sec (x) + tan (x) ∣

\frac{d}{d x} (\frac{4 sin ( x )}{3 + cos ( x )})

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