What is the integral from 0 to 3 of x^3e^{x^4} ?

The integral from 0 to 3 of x^3e^{x^4} is (e^{81}-1)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of x^3e^{x^4}

\int_{0}^{3} x^{3} e^{x^{4}} d x

\frac{e ^{81} - 1}{4}

Decimal

3.76524 E 34

Solution steps

\int_{0}^{3} x^{3} e^{x^{4}} d x

Apply u-substitution

: \int_{0}^{27} \frac{e ^{u^{\frac{4}{3}}} u ^{\frac{1}{3}}}{3} d u

= \int_{0}^{27} \frac{e ^{u^{\frac{4}{3}}} u ^{\frac{1}{3}}}{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^{\frac{4}{3}}} u^{\frac{1}{3}} d u

Apply u-substitution

: \int_{0}^{81} \frac{3 e ^{v}}{4} d v

= \frac{1}{3} \cdot \int_{0}^{81} \frac{3 e ^{v}}{4} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \frac{3}{4} \cdot \int_{0}^{81} e^{v} d v

Use the common integral:

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

= \frac{1}{3} \cdot \frac{3}{4} [e^{v}]_{0}^{81}

Simplify

\frac{1}{3} \cdot \frac{3}{4} [e^{v}]_{0}^{81} : \frac{1}{4} [e^{v}]_{0}^{81}

= \frac{1}{4} [e^{v}]_{0}^{81}

Compute the boundaries:

e^{81} - 1

= \frac{1}{4} (e^{81} - 1)

Simplify

= \frac{e ^{81} - 1}{4}

\int_{0}^{1} (4 x^{2} + x) d x

\int_{0}^{10} \frac{24 x}{2 x + 3} d x

\int_{2}^{4} x 16 - x^{2} d x

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

\int_{- 1}^{4} \frac{x}{x + 5} d x

What is the integral from 0 to 3 of x^3e^{x^4} ?
The integral from 0 to 3 of x^3e^{x^4} is (e^{81}-1)/4