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

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

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

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

3

Solution steps

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

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 4 x d x = 30

\int_{1}^{4} 4 d x = 12

= 21 - 30 + 12

Simplify

= 3

\int_{x}^{x^{2}} (e^{x^{4}}) d x

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

\int_{0}^{5} \frac{x}{5} d x

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

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

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