What is the integral from 0 to 1 of y^3e^{y^4} ?

The integral from 0 to 1 of y^3e^{y^4} is (e-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 1 of y^3e^{y^4}

\int_{0}^{1} y^{3} e^{y^{4}} d y

\frac{e - 1}{4}

Decimal Notation

0.42957 \dots

Solution steps

\int_{0}^{1} y^{3} e^{y^{4}} d y

Apply u-substitution

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

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

Apply u-substitution

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

= \frac{1}{3} \cdot \int_{0}^{1} \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}^{1} 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}^{1}

Simplify

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

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

Compute the boundaries:

e - 1

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

Simplify

= \frac{e - 1}{4}

\int_{- 1}^{1} c x^{2} d x

\int_{- 1}^{0} 2 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

\int_{- 1}^{1} - 4 x^{2} + 3 x - 2 d x

\int_{1}^{e} (23 + ln (x))^{2} d x

\int_{π}^{2 π} sin^{3} (x) d x

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