What is the integral from-3 to 0 of x^2e^{x^3+27} ?

The integral from-3 to 0 of x^2e^{x^3+27} is (e^{27}-1)/3

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integral from-3 to 0 of x^2e^{x^3+27}

\int_{- 3}^{0} x^{2} e^{x^{3} + 27} d x

\frac{e ^{27} - 1}{3}

Decimal

177349413533.59927

Solution steps

\int_{- 3}^{0} x^{2} e^{x^{3} + 27} d x

Apply u-substitution

: \int_{0}^{27} \frac{e ^{u}}{3} d u

= \int_{0}^{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}{3} \cdot \int_{0}^{27} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{27}

Compute the boundaries:

e^{27} - 1

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

Simplify

= \frac{e ^{27} - 1}{3}

\frac{d}{d z} (z^{2} e^{z})

f (x) = \frac{1}{( 1 + x )}

\frac{d}{d x} (4 ln ∣ x - e^{x} ∣)

y = \frac{cos ( x )}{1 - sin ( x )}

y = cos (sin (tan (9 x)))

What is the integral from-3 to 0 of x^2e^{x^3+27} ?
The integral from-3 to 0 of x^2e^{x^3+27} is (e^{27}-1)/3