What is the integral from 0 to 1 of (6x^2-1)e^x ?

The integral from 0 to 1 of (6x^2-1)e^x is 5e-11

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 (6x^2-1)e^x

\int_{0}^{1} (6 x^{2} - 1) e^{x} d x

5 e - 11

Decimal Notation

2.59140 \dots

Solution steps

\int_{0}^{1} (6 x^{2} - 1) e^{x} d x

Apply Integration By Parts

: [(6 x^{2} - 1) e^{x} - \int 12 x e^{x} d x]_{0}^{1}

= [(6 x^{2} - 1) e^{x} - \int 12 x e^{x} d x]_{0}^{1}

\int 12 x e^{x} d x = 12 (e^{x} x - e^{x})

= [(6 x^{2} - 1) e^{x} - 12 (e^{x} x - e^{x})]_{0}^{1}

Expand

(6 x^{2} - 1) e^{x} - 12 (e^{x} x - e^{x}) : 6 e^{x} x^{2} - 12 e^{x} x + 11 e^{x}

= [6 e^{x} x^{2} - 12 e^{x} x + 11 e^{x}]_{0}^{1}

Compute the boundaries:

5 e - 11

= 5 e - 11

x \to - 1 lim (\frac{1}{x ^{2}} + 3 x - 1)

\frac{d y}{d x} = (y + 1)^{(- 1)}

x^{^{''}} + 25 x = 0

\frac{d}{d x} (e^{x} cos (x))

\frac{d}{d x} (ln (e^{x} + 7 x + 1))

What is the integral from 0 to 1 of (6x^2-1)e^x ?
The integral from 0 to 1 of (6x^2-1)e^x is 5e-11