What is the integral from 2 to 6 of (x^2+7x-5)(x-2) ?

The integral from 2 to 6 of (x^2+7x-5)(x-2) is 1208/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 2 to 6 of (x^2+7x-5)(x-2)

\int_{2}^{6} (x^{2} + 7 x - 5) (x - 2) d x

\frac{1208}{3}

Decimal

402.66666 \dots

Solution steps

\int_{2}^{6} (x^{2} + 7 x - 5) (x - 2) d x

Expand

(x^{2} + 7 x - 5) (x - 2) : x^{3} + 5 x^{2} - 19 x + 10

= \int_{2}^{6} x^{3} + 5 x^{2} - 19 x + 10 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_{2}^{6} x^{3} d x + \int_{2}^{6} 5 x^{2} d x - \int_{2}^{6} 19 x d x + \int_{2}^{6} 10 d x

\int_{2}^{6} x^{3} d x = 320

\int_{2}^{6} 5 x^{2} d x = \frac{1040}{3}

\int_{2}^{6} 19 x d x = 304

\int_{2}^{6} 10 d x = 40

= 320 + \frac{1040}{3} - 304 + 40

Simplify

320 + \frac{1040}{3} - 304 + 40 : \frac{1208}{3}

= \frac{1208}{3}

What is the integral from 2 to 6 of (x^2+7x-5)(x-2) ?
The integral from 2 to 6 of (x^2+7x-5)(x-2) is 1208/3