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

The integral from 0 to 1 of (4x-3)^2 is 7/3

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integral from 0 to 1 of (4x-3)^2

\int_{0}^{1} (4 x - 3)^{2} d x

\frac{7}{3}

Decimal

2.33333 \dots

Solution steps

\int_{0}^{1} (4 x - 3)^{2} d x

Expand

(4 x - 3)^{2} : 16 x^{2} - 24 x + 9

= \int_{0}^{1} 16 x^{2} - 24 x + 9 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} 16 x^{2} d x - \int_{0}^{1} 24 x d x + \int_{0}^{1} 9 d x

\int_{0}^{1} 16 x^{2} d x = \frac{16}{3}

\int_{0}^{1} 24 x d x = 12

\int_{0}^{1} 9 d x = 9

= \frac{16}{3} - 12 + 9

Simplify

\frac{16}{3} - 12 + 9 : \frac{7}{3}

= \frac{7}{3}

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What is the integral from 0 to 1 of (4x-3)^2 ?
The integral from 0 to 1 of (4x-3)^2 is 7/3