What is the integral of-0.08x^2+4x ?

The integral of-0.08x^2+4x is -0.02666…x^3+2x^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-0.08x^2+4x

\int - 0.08 x^{2} + 4 x d x

- 0.02666 \dots x^{3} + 2 x^{2} + C

Solution steps

\int - 0.08 x^{2} + 4 x 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 0.08 x^{2} d x + \int 4 x d x

\int 0.08 x^{2} d x = 0.02666 \dots x^{3}

\int 4 x d x = 2 x^{2}

= - 0.02666 \dots x^{3} + 2 x^{2}

Add a constant to the solution

= - 0.02666 \dots x^{3} + 2 x^{2} + C

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

\int (x^{2} + x) (e^{1 - 2 x}) d x

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

\int cot^{3} (y) csc^{2} (y) d y

What is the integral of-0.08x^2+4x ?
The integral of-0.08x^2+4x is -0.02666…x^3+2x^2+C