What is the integral from 0 to 5 of (2x-1)^2 ?

The integral from 0 to 5 of (2x-1)^2 is 365/3

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integral from 0 to 5 of (2x-1)^2

\int_{0}^{5} (2 x - 1)^{2} d x

\frac{365}{3}

Decimal

121.66666 \dots

Solution steps

\int_{0}^{5} (2 x - 1)^{2} d x

Expand

(2 x - 1)^{2} : 4 x^{2} - 4 x + 1

= \int_{0}^{5} 4 x^{2} - 4 x + 1 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}^{5} 4 x^{2} d x - \int_{0}^{5} 4 x d x + \int_{0}^{5} 1 d x

\int_{0}^{5} 4 x^{2} d x = \frac{500}{3}

\int_{0}^{5} 4 x d x = 50

\int_{0}^{5} 1 d x = 5

= \frac{500}{3} - 50 + 5

Simplify

\frac{500}{3} - 50 + 5 : \frac{365}{3}

= \frac{365}{3}

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What is the integral from 0 to 5 of (2x-1)^2 ?
The integral from 0 to 5 of (2x-1)^2 is 365/3