What is the integral from-2 to 1 of x^2e^{x^3+8} ?

The integral from-2 to 1 of x^2e^{x^3+8} is (e^9-1)/3

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integral from-2 to 1 of x^2e^{x^3+8}

\int_{- 2}^{1} x^{2} e^{x^{3} + 8} d x

\frac{e ^{9} - 1}{3}

Decimal

2700.69464 \dots

Solution steps

\int_{- 2}^{1} x^{2} e^{x^{3} + 8} d x

Apply u-substitution

: \int_{0}^{9} \frac{e ^{u}}{3} d u

= \int_{0}^{9} \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}^{9} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{9}

Compute the boundaries:

e^{9} - 1

= \frac{1}{3} (e^{9} - 1)

Simplify

= \frac{e ^{9} - 1}{3}

y^{^{'}} + 3 y = 1 - 6 e^{- 2 x}

\int x^{2} \cdot 6^{x^{3}} d x

x \to - 9 + lim (\frac{∣ x + 9 ∣}{x + 9})

\int \frac{2 x}{x + 1} d x

h \to 0 lim (\frac{( 1 + h ) ^{6} - 1}{h})

What is the integral from-2 to 1 of x^2e^{x^3+8} ?
The integral from-2 to 1 of x^2e^{x^3+8} is (e^9-1)/3