What is the integral from 0 to 3 of (3x^4-xy+y^3) ?

The integral from 0 to 3 of (3x^4-xy+y^3) is 81/4+9x^4-(9x)/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 (3x^4-xy+y^3)

\int_{0}^{3} (3 x^{4} - x y + y^{3}) d y

\frac{81}{4} + 9 x^{4} - \frac{9 x}{2}

Solution steps

\int_{0}^{3} 3 x^{4} - x y + y^{3} d y

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} y^{3} d y + \int_{0}^{3} 3 x^{4} d y - \int_{0}^{3} x y d y

\int_{0}^{3} y^{3} d y = \frac{81}{4}

\int_{0}^{3} 3 x^{4} d y = 9 x^{4}

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

= \frac{81}{4} + 9 x^{4} - \frac{9 x}{2}

\int_{\frac{π}{6}}^{\frac{π}{2}} sin (2 x) d x

\int_{- 1}^{2} x - 8 ∣ x ∣ d x

\int_{- 9}^{9} x 81 - x^{2} d x

\int_{0}^{s i n (x)} sin (t^{2}) d t

What is the integral from 0 to 3 of (3x^4-xy+y^3) ?
The integral from 0 to 3 of (3x^4-xy+y^3) is 81/4+9x^4-(9x)/2