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

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

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

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Take the constant out:

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

= \frac{1}{16} \cdot \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

= \frac{1}{16} (\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{1}{16} (\frac{64}{3} - 64 + 64)

Simplify

\frac{1}{16} (\frac{64}{3} - 64 + 64) : \frac{4}{3}

= \frac{4}{3}

\int_{0}^{2} 0.5 x^{2} d x

\int_{0}^{1} x \frac{2 ( x + 1 )}{3} d x

\int_{1}^{1.5} \frac{3}{8} x^{2} d x

\int_{- 2}^{- 1} \frac{3}{y} d y

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