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^4-(sqrt(pi))/4 \erfi(2)-e/2+(sqrt(pi))/4 \erfi(1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 2 of x^2e^{x^{3-1}}

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

e^{4} - \frac{π}{4} erfi (2) - \frac{e}{2} + \frac{π}{4} erfi (1)

Solution steps

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

Simplify

= \int_{1}^{2} e^{x^{2}} x^{2} d x

Simplify

x^{2} : xx

= \int_{1}^{2} e^{x^{2}} xx d x

Apply Integration By Parts

: [\frac{1}{2} e^{x^{2}} x - \int \frac{1}{2} e^{x^{2}} d x]_{1}^{2}

= [\frac{1}{2} e^{x^{2}} x - \int \frac{1}{2} e^{x^{2}} d x]_{1}^{2}

\int \frac{1}{2} e^{x^{2}} d x = \frac{π}{4} erfi (x)

= [\frac{1}{2} e^{x^{2}} x - \frac{π}{4} erfi (x)]_{1}^{2}

Compute the boundaries:

e^{4} - \frac{π}{4} erfi (2) - \frac{e}{2} + \frac{π}{4} erfi (1)

= e^{4} - \frac{π}{4} erfi (2) - \frac{e}{2} + \frac{π}{4} erfi (1)

\int_{0}^{1} 3 x (x^{2} - 1)^{13} d x

\int_{x^{2}}^{2 x} 4 x y d y

\int_{0}^{\frac{π}{4}} sec^{4} (θ) d θ

\int_{0}^{4} x^{3} - 8 x^{2} + 16 x d x

\int_{1}^{3} t^{2} + 3 t - 18 d t

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^4-(sqrt(pi))/4 \erfi(2)-e/2+(sqrt(pi))/4 \erfi(1)