What is the integral of x^2-16x+64 ?

The integral of x^2-16x+64 is (x^3)/3-8x^2+64x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^2-16x+64

\int x^{2} - 16 x + 64 d x

\frac{x ^{3}}{3} - 8 x^{2} + 64 x + C

Solution steps

\int x^{2} - 16 x + 64 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 x^{2} d x - \int 16 x d x + \int 64 d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 16 x d x = 8 x^{2}

\int 64 d x = 64 x

= \frac{x ^{3}}{3} - 8 x^{2} + 64 x

Add a constant to the solution

= \frac{x ^{3}}{3} - 8 x^{2} + 64 x + C

\int_{0.1}^{0.2} 400 x d x

\int_{0}^{l n (3)} e^{x} - e^{- 2 x} d x

\frac{d y}{d x} + \frac{2}{x} y = 6 x^{4} y^{2}

\frac{d y}{d x} = \frac{y}{x + 2 y}

4 e^{x + 4} + 1

What is the integral of x^2-16x+64 ?
The integral of x^2-16x+64 is (x^3)/3-8x^2+64x+C