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

The integral from 1 to 4 of (x-(x-2)^2) is 9/2

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

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

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

Expand

x - (x - 2)^{2} : - x^{2} + 5 x - 4

= \int_{1}^{4} - x^{2} + 5 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} 5 x d x - \int_{1}^{4} 4 d x

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

\int_{1}^{4} 5 x d x = \frac{75}{2}

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

= - 21 + \frac{75}{2} - 12

Simplify

- 21 + \frac{75}{2} - 12 : \frac{9}{2}

= \frac{9}{2}

\int_{\frac{π}{4}}^{\frac{3 π}{4}} 1

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\int_{π}^{2 π} sin (t) sin (n t) d t

\int_{- 1}^{1} 4 x^{2} + 7 d x

\int_{0}^{4} 6 e^{2 t} d t

What is the integral from 1 to 4 of (x-(x-2)^2) ?
The integral from 1 to 4 of (x-(x-2)^2) is 9/2