What is the integral from x^3 to 1 of 9x^2y ?

The integral from x^3 to 1 of 9x^2y is (9x^2(1-x^6))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from x^3 to 1 of 9x^2y

\int_{x^{3}}^{1} 9 x^{2} y d y

\frac{9 x ^{2} ( 1 - x ^{6} )}{2}

Solution steps

\int_{x^{3}}^{1} 9 x^{2} y d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 x^{2} \cdot \int_{x^{3}}^{1} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{x^{3}}^{1}

= 9 x^{2} [\frac{y ^{2}}{2}]_{x^{3}}^{1}

Compute the boundaries:

\frac{1}{2} - \frac{x ^{6}}{2}

= 9 x^{2} (\frac{1}{2} - \frac{x ^{6}}{2})

Simplify

= \frac{9 x ^{2} ( 1 - x ^{6} )}{2}

\int_{0}^{π} (sin (x) - 7) d x

\int_{0}^{5} (\frac{50}{x + 5} - 5) d x

\int_{1}^{3} (x - 2)^{4} d x

\int_{0}^{1} e^{y} y d y

\int_{- 3}^{3} (x^{\frac{1}{3}}) d x

What is the integral from x^3 to 1 of 9x^2y ?
The integral from x^3 to 1 of 9x^2y is (9x^2(1-x^6))/2