What is the integral from-3 to 1 of x^2e^{x^3+27} ?

The integral from-3 to 1 of x^2e^{x^3+27} is (e^{28}-1)/3

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integral from-3 to 1 of x^2e^{x^3+27}

\int_{- 3}^{1} x^{2} e^{x^{3} + 27} d x

\frac{e ^{28} - 1}{3}

Decimal Notation

482085688096.8243

Solution steps

\int_{- 3}^{1} x^{2} e^{x^{3} + 27} d x

Apply u-substitution

: \int_{0}^{28} \frac{e ^{u}}{3} d u

= \int_{0}^{28} \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}^{28} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{28}

Compute the boundaries:

e^{28} - 1

= \frac{1}{3} (e^{28} - 1)

Simplify

= \frac{e ^{28} - 1}{3}

\frac{( 3 s + 3 )}{s ^{2} - 9}

\int cos^{3} (6 x) d x

x + y = 0, x + y^{2} = 56

x^{n}

\frac{d}{d x} (x^{7} x)

What is the integral from-3 to 1 of x^2e^{x^3+27} ?
The integral from-3 to 1 of x^2e^{x^3+27} is (e^{28}-1)/3