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

The integral from 1 to 2 of (2x+1)^2 is 49/3

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

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

\frac{49}{3}

Decimal

16.33333 \dots

Solution steps

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

Expand

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

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

\int_{1}^{2} 4 x^{2} d x = \frac{28}{3}

\int_{1}^{2} 4 x d x = 6

\int_{1}^{2} 1 d x = 1

= \frac{28}{3} + 6 + 1

Simplify

\frac{28}{3} + 6 + 1 : \frac{49}{3}

= \frac{49}{3}

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What is the integral from 1 to 2 of (2x+1)^2 ?
The integral from 1 to 2 of (2x+1)^2 is 49/3