What is the integral from 0 to 1 of (x^2-4x+4) ?

The integral from 0 to 1 of (x^2-4x+4) is 7/3

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integral from 0 to 1 of (x^2-4x+4)

\int_{0}^{1} (x^{2} - 4 x + 4) d x

\frac{7}{3}

Decimal

2.33333 \dots

Solution steps

\int_{0}^{1} x^{2} - 4 x + 4 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}^{1} x^{2} d x - \int_{0}^{1} 4 x d x + \int_{0}^{1} 4 d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} 4 x d x = 2

\int_{0}^{1} 4 d x = 4

= \frac{1}{3} - 2 + 4

Simplify

\frac{1}{3} - 2 + 4 : \frac{7}{3}

= \frac{7}{3}

\int_{0}^{u} 11 + 12 t^{2} d t

\int_{4}^{5} x^{2} - 4 x d x

\int_{0}^{9} (x - \frac{1}{2}) d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} 2 x cos (x) d x

\int_{0}^{1} 4 x - 4 d x

What is the integral from 0 to 1 of (x^2-4x+4) ?
The integral from 0 to 1 of (x^2-4x+4) is 7/3