What is the integral from-2 to x of (t^2-2t) ?

The integral from-2 to x of (t^2-2t) is (x^3+8)/3-x^2+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-2 to x of (t^2-2t)

\int_{- 2}^{x} (t^{2} - 2 t) d t

\frac{x ^{3} + 8}{3} - x^{2} + 4

Solution steps

\int_{- 2}^{x} t^{2} - 2 t d t

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}^{x} t^{2} d t - \int_{- 2}^{x} 2 t d t

\int_{- 2}^{x} t^{2} d t = \frac{x ^{3} + 8}{3}

\int_{- 2}^{x} 2 t d t = x^{2} - 4

= \frac{x ^{3} + 8}{3} - (x^{2} - 4)

Simplify

= \frac{x ^{3} + 8}{3} - x^{2} + 4

\int_{3}^{7} \frac{10}{x ^{2}} d x

\int_{0}^{2} \frac{2 x}{3 - x} d x

\int_{1}^{25} \frac{11}{x} d x

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

\int_{0}^{a} 8 x a^{2} - x^{2} d x

What is the integral from-2 to x of (t^2-2t) ?
The integral from-2 to x of (t^2-2t) is (x^3+8)/3-x^2+4