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

The integral from-2 to 2 of x^2e^{x^3+8} is (e^{16}-1)/3

Popular >

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

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

\frac{e ^{16} - 1}{3}

Decimal Notation

2962036.50683 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{16} \frac{e ^{u}}{3} d u

= \int_{0}^{16} \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}^{16} e^{u} d u

Use the common integral:

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

= \frac{1}{3} [e^{u}]_{0}^{16}

Compute the boundaries:

e^{16} - 1

= \frac{1}{3} (e^{16} - 1)

Simplify

= \frac{e ^{16} - 1}{3}

5 ln (x)

x^{2} lo g_{3} (x^{3})

\int (8 x - \frac{7}{x}) d x

y = cos (x), at x = \frac{3 π}{2}

f (x) = \frac{5}{( 2 x - 3 ) ^{4}}

What is the integral from-2 to 2 of x^2e^{x^3+8} ?
The integral from-2 to 2 of x^2e^{x^3+8} is (e^{16}-1)/3