What is the integral from 0 to 4 of x^2-8x+16 ?

The integral from 0 to 4 of x^2-8x+16 is 64/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of x^2-8x+16

\int_{0}^{4} x^{2} - 8 x + 16 d x

\frac{64}{3}

Decimal

21.33333 \dots

Solution steps

\int_{0}^{4} x^{2} - 8 x + 16 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}^{4} x^{2} d x - \int_{0}^{4} 8 x d x + \int_{0}^{4} 16 d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 8 x d x = 64

\int_{0}^{4} 16 d x = 64

= \frac{64}{3} - 64 + 64

Simplify

= \frac{64}{3}

\int_{0}^{π} a sin (x) d x

\int_{4}^{2} - 2 d x

\int_{4}^{9} \frac{2 + t}{t} d t

\int_{2}^{7} 5 arccot (x) d x

\int_{0}^{3} 5^{x} ln (5) d x

What is the integral from 0 to 4 of x^2-8x+16 ?
The integral from 0 to 4 of x^2-8x+16 is 64/3