What is the integral from 0 to 1 of x^3e^{sqrt(x)} ?

The integral from 0 to 1 of x^3e^{sqrt(x)} is 2(-1854e+5040)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of x^3e^{sqrt(x)}

\int_{0}^{1} x^{3} e^{x} d x

2 (- 1854 e + 5040)

Decimal

0.61098 \dots

Solution steps

\int_{0}^{1} x^{3} e^{x} d x

Apply u-substitution

Take the constant out:

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

Apply u-substitution

= \frac{1}{3} \cdot \int_{0}^{1} 6 e^{v} v^{7} d v

Take the constant out:

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

= \frac{1}{3} \cdot 6 \cdot \int_{0}^{1} e^{v} v^{7} d v

Apply u-substitution

= \frac{1}{3} \cdot 6 \cdot \int_{1}^{e} ln^{7} (w) d w

Apply Integration By Parts

= \frac{1}{3} \cdot 6 [w ln^{7} (w) - \int 7 ln^{6} (w) d w]_{1}^{e}

\int 7 ln^{6} (w) d w = 7 (w ln^{6} (w) - 6 (w ln^{5} (w) - 5 (w ln^{4} (w) - 4 (w ln^{3} (w) - 3 (w ln^{2} (w) - 2 (w ln (w) - w))))))

= \frac{1}{3} \cdot 6 [w ln^{7} (w) - 7 (w ln^{6} (w) - 6 (w ln^{5} (w) - 5 (w ln^{4} (w) - 4 (w ln^{3} (w) - 3 (w ln^{2} (w) - 2 (w ln (w) - w))))))]_{1}^{e}

Simplify

\frac{1}{3} \cdot 6 [w ln^{7} (w) - 7 (w ln^{6} (w) - 6 (w ln^{5} (w) - 5 (w ln^{4} (w) - 4 (w ln^{3} (w) - 3 (w ln^{2} (w) - 2 (w ln (w) - w))))))]_{1}^{e} : 2 [w ln^{7} (w) - 7 (w ln^{6} (w) - 6 (w ln^{5} (w) - 5 (w ln^{4} (w) - 4 (w ln^{3} (w) - 3 (w ln^{2} (w) - 2 (w ln (w) - w))))))]_{1}^{e}

= 2 [w ln^{7} (w) - 7 (w ln^{6} (w) - 6 (w ln^{5} (w) - 5 (w ln^{4} (w) - 4 (w ln^{3} (w) - 3 (w ln^{2} (w) - 2 (w ln (w) - w))))))]_{1}^{e}

Compute the boundaries:

- 1854 e + 5040

= 2 (- 1854 e + 5040)

What is the integral from 0 to 1 of x^3e^{sqrt(x)} ?
The integral from 0 to 1 of x^3e^{sqrt(x)} is 2(-1854e+5040)