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^9-1)/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 ^{9} - 1}{3}

Decimal Notation

2700.69464 \dots

Solution steps

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

f (x) = ln (sec (x))

\int_{1}^{x} t^{4} d t

f (x) = (x^{2} - 2)^{2}

\frac{d}{d x} (ln (3 x^{2} + 2 x))

x \to 0 + lim (csc (x) - cot (x))

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^9-1)/3