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

The integral from 1 to 2 of x^2e^{x^3-1} is (e^7-1)/3

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

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

\frac{e ^{7} - 1}{3}

Decimal Notation

365.21105 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{7} \frac{e ^{u}}{3} d u

= \int_{0}^{7} \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}^{7} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{7}

Compute the boundaries:

e^{7} - 1

= \frac{1}{3} (e^{7} - 1)

Simplify

= \frac{e ^{7} - 1}{3}

y^{^{''}} + 9 y = (sec (3 t))^{2}

\frac{d y}{d x} + \frac{2 y}{x} = \frac{sin ( x )}{x ^{2}}

t e^{- t}

\frac{d}{d x} (e^{- 2 x} - 3 y)

What is the integral from 1 to 2 of x^2e^{x^3-1} ?
The integral from 1 to 2 of x^2e^{x^3-1} is (e^7-1)/3