What is the integral from 2 to 3 of x^2e^{x^3-8} ?

The integral from 2 to 3 of x^2e^{x^3-8} is (e^{27}-e^8)/(3e^8)

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integral from 2 to 3 of x^2e^{x^3-8}

\int_{2}^{3} x^{2} e^{x^{3} - 8} d x

\frac{e ^{27} - e ^{8}}{3 e ^{8}}

Decimal Notation

59494099.98772 \dots

Solution steps

\int_{2}^{3} x^{2} e^{x^{3} - 8} d x

x^{2} e^{x^{3} - 8} = e^{- 8} x^{2} e^{x^{3}}

= \int_{2}^{3} e^{- 8} x^{2} e^{x^{3}} d x

Take the constant out:

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

= e^{- 8} \cdot \int_{2}^{3} x^{2} e^{x^{3}} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{8}} \cdot \int_{2}^{3} x^{2} e^{x^{3}} d x

Apply u-substitution

: \int_{8}^{27} \frac{e ^{u}}{3} d u

= \frac{1}{e ^{8}} \cdot \int_{8}^{27} \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}{e ^{8}} \cdot \frac{1}{3} \cdot \int_{8}^{27} e^{u} d u

Use the common integral:

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

= \frac{1}{e ^{8}} \cdot \frac{1}{3} [e^{u}]_{8}^{27}

Simplify

\frac{1}{e ^{8}} \cdot \frac{1}{3} [e^{u}]_{8}^{27} : \frac{1}{3 e ^{8}} [e^{u}]_{8}^{27}

= \frac{1}{3 e ^{8}} [e^{u}]_{8}^{27}

Compute the boundaries:

e^{27} - e^{8}

= \frac{1}{3 e ^{8}} (e^{27} - e^{8})

Simplify

= \frac{e ^{27} - e ^{8}}{3 e ^{8}}

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What is the integral from 2 to 3 of x^2e^{x^3-8} ?
The integral from 2 to 3 of x^2e^{x^3-8} is (e^{27}-e^8)/(3e^8)