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

The integral from 0 to 4 of (4x-1)^2 is 844/3

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

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

\frac{844}{3}

Decimal

281.33333 \dots

Solution steps

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

Expand

(4 x - 1)^{2} : 16 x^{2} - 8 x + 1

= \int_{0}^{4} 16 x^{2} - 8 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}^{4} 16 x^{2} d x - \int_{0}^{4} 8 x d x + \int_{0}^{4} 1 d x

\int_{0}^{4} 16 x^{2} d x = \frac{1024}{3}

\int_{0}^{4} 8 x d x = 64

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

= \frac{1024}{3} - 64 + 4

Simplify

\frac{1024}{3} - 64 + 4 : \frac{844}{3}

= \frac{844}{3}

\int_{0}^{8} \frac{1}{2 x ^{\frac{1}{3}}} d x

\int_{0}^{2} (- x^{4} + 2 x^{2} + 8) d x

\int_{0}^{5} 3 x - 8 d x

\int_{0}^{1} t e^{- t^{2}} d t

\int_{0}^{2} 13 x (x^{2} + 1)^{2} d x

What is the integral from 0 to 4 of (4x-1)^2 ?
The integral from 0 to 4 of (4x-1)^2 is 844/3