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

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

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

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

\frac{7}{3}

Decimal

2.33333 \dots

Solution steps

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

Expand

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

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

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

\int_{0}^{1} 8 x d x = 4

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

= \frac{16}{3} - 4 + 1

Simplify

\frac{16}{3} - 4 + 1 : \frac{7}{3}

= \frac{7}{3}

\int_{0}^{3} 1 + \frac{81}{64} x d x

\int_{0}^{2 π} - 16 sin^{2} (x) d x

\int_{4}^{4} x d x

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

\int_{0}^{1} a x^{2} + b x^{3} d x

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