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

The integral from 0 to 1 of (2t-1)^2 is 1/3

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integral from 0 to 1 of (2t-1)^2

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

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

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

Expand

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

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

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} 4 t^{2} d t - \int_{0}^{1} 4 t d t + \int_{0}^{1} 1 d t

\int_{0}^{1} 4 t^{2} d t = \frac{4}{3}

\int_{0}^{1} 4 t d t = 2

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

= \frac{4}{3} - 2 + 1

Simplify

\frac{4}{3} - 2 + 1 : \frac{1}{3}

= \frac{1}{3}

\frac{d}{d x} (sin^{2} (5 x - 3))

\frac{d y}{d x} + \frac{5}{x} \cdot y = 0

- 8 x^{- 1} + 4 x^{- 2} - 7

n = 0 \sum \infty 1 2^{n} x^{n}

\frac{\partial}{\partial y} ((x - 3)^{2} + (y - 4)^{2})

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