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

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

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

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

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

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

Expand

(2 x + 1)^{2} : 4 x^{2} + 4 x + 1

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

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

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

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

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

Simplify

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

= \frac{1}{3}

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