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

The integral from-1 to 0 of (2w+3)^2 is 13/3

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integral from-1 to 0 of (2w+3)^2

\int_{- 1}^{0} (2 w + 3)^{2} d w

\frac{13}{3}

Decimal Notation

4.33333 \dots

Solution steps

\int_{- 1}^{0} (2 w + 3)^{2} d w

Expand

(2 w + 3)^{2} : 4 w^{2} + 12 w + 9

= \int_{- 1}^{0} 4 w^{2} + 12 w + 9 d w

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 w^{2} d w + \int_{- 1}^{0} 12 w d w + \int_{- 1}^{0} 9 d w

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

\int_{- 1}^{0} 12 w d w = - 6

\int_{- 1}^{0} 9 d w = 9

= \frac{4}{3} - 6 + 9

Simplify

\frac{4}{3} - 6 + 9 : \frac{13}{3}

= \frac{13}{3}

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What is the integral from-1 to 0 of (2w+3)^2 ?
The integral from-1 to 0 of (2w+3)^2 is 13/3