What is the integral from x^2 to 0 of (t-t^2) ?

The integral from x^2 to 0 of (t-t^2) is -(x^4)/2+(x^6)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from x^2 to 0 of (t-t^2)

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

- \frac{x ^{4}}{2} + \frac{x ^{6}}{3}

Solution steps

\int_{x^{2}}^{0} t - t^{2} 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_{x^{2}}^{0} t d t - \int_{x^{2}}^{0} t^{2} d t

\int_{x^{2}}^{0} t d t = - \frac{x ^{4}}{2}

\int_{x^{2}}^{0} t^{2} d t = - \frac{x ^{6}}{3}

= - \frac{x ^{4}}{2} - (- \frac{x ^{6}}{3})

Simplify

= - \frac{x ^{4}}{2} + \frac{x ^{6}}{3}

What is the integral from x^2 to 0 of (t-t^2) ?
The integral from x^2 to 0 of (t-t^2) is -(x^4)/2+(x^6)/3