What is the integral from 0 to 1 of (8x-1)^2 ?

The integral from 0 to 1 of (8x-1)^2 is 43/3

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integral from 0 to 1 of (8x-1)^2

\int_{0}^{1} (8 x - 1)^{2} d x

\frac{43}{3}

Decimal Notation

14.33333 \dots

Solution steps

\int_{0}^{1} (8 x - 1)^{2} d x

Expand

(8 x - 1)^{2} : 64 x^{2} - 16 x + 1

= \int_{0}^{1} 64 x^{2} - 16 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}^{1} 64 x^{2} d x - \int_{0}^{1} 16 x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} 64 x^{2} d x = \frac{64}{3}

\int_{0}^{1} 16 x d x = 8

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

= \frac{64}{3} - 8 + 1

Simplify

\frac{64}{3} - 8 + 1 : \frac{43}{3}

= \frac{43}{3}

\int_{0}^{1} 7^{3 x} d x

\int \frac{( ln ( x ) - 1 ) ^{2}}{x} d x

y = \frac{x}{3} + \frac{3 x ^{2}}{2}

What is the integral from 0 to 1 of (8x-1)^2 ?
The integral from 0 to 1 of (8x-1)^2 is 43/3