What is the integral from 3 to 4 of (x^2-x-6) ?

The integral from 3 to 4 of (x^2-x-6) is 17/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 3 to 4 of (x^2-x-6)

\int_{3}^{4} (x^{2} - x - 6) d x

\frac{17}{6}

Decimal

2.83333 \dots

Solution steps

\int_{3}^{4} x^{2} - x - 6 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_{3}^{4} x^{2} d x - \int_{3}^{4} x d x - \int_{3}^{4} 6 d x

\int_{3}^{4} x^{2} d x = \frac{37}{3}

\int_{3}^{4} x d x = \frac{7}{2}

\int_{3}^{4} 6 d x = 6

= \frac{37}{3} - \frac{7}{2} - 6

Simplify

\frac{37}{3} - \frac{7}{2} - 6 : \frac{17}{6}

= \frac{17}{6}

\int_{2 x}^{10} cos (y^{2}) d y

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (u) d u

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

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

\int_{0}^{2} x (4 - x^{2}) d x

What is the integral from 3 to 4 of (x^2-x-6) ?
The integral from 3 to 4 of (x^2-x-6) is 17/6