What is the integral from-1 to 3 of (2x^2-8) ?

The integral from-1 to 3 of (2x^2-8) is -40/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 3 of (2x^2-8)

\int_{- 1}^{3} (2 x^{2} - 8) d x

- \frac{40}{3}

Decimal Notation

- 13.33333 \dots

Solution steps

\int_{- 1}^{3} 2 x^{2} - 8 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{3} 2 x^{2} d x - \int_{- 1}^{3} 8 d x

\int_{- 1}^{3} 2 x^{2} d x = \frac{56}{3}

\int_{- 1}^{3} 8 d x = 32

= \frac{56}{3} - 32

Simplify

\frac{56}{3} - 32 : - \frac{40}{3}

= - \frac{40}{3}

\frac{d}{d x} (- \frac{18}{a x})

\frac{\partial}{\partial x} (4 x sin (3 x^{2} y))

\int_{- 2}^{18} \frac{1}{∣ 2 x ∣} d x

\frac{d}{d x} ((4 x^{3} + 6) (x + 8))

x \to \infty lim (7 x^{2} + 8)

What is the integral from-1 to 3 of (2x^2-8) ?
The integral from-1 to 3 of (2x^2-8) is -40/3