What is the integral from 0 to 3 of (-x^2+x-1) ?

The integral from 0 to 3 of (-x^2+x-1) is -15/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of (-x^2+x-1)

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

- \frac{15}{2}

Decimal Notation

- 7.5

Solution steps

\int_{0}^{3} - x^{2} + x - 1 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_{0}^{3} x^{2} d x + \int_{0}^{3} x d x - \int_{0}^{3} 1 d x

\int_{0}^{3} x^{2} d x = 9

\int_{0}^{3} x d x = \frac{9}{2}

\int_{0}^{3} 1 d x = 3

= - 9 + \frac{9}{2} - 3

Simplify

- 9 + \frac{9}{2} - 3 : - \frac{15}{2}

= - \frac{15}{2}

\int_{1}^{16} \frac{1}{x} d x

\int \frac{2 x}{3 ( x ^{2} + 1 )} d x

\frac{d}{d x} (x^{12} ln (x))

\int \frac{16}{x ( x ^{2} + 4 ) ^{2}} d x

\frac{d}{d x} ((x sin (2 x))^{\frac{3}{2}})

What is the integral from 0 to 3 of (-x^2+x-1) ?
The integral from 0 to 3 of (-x^2+x-1) is -15/2