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

The integral from-1 to 4 of (2x+4)^2 is 860/3

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

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

\frac{860}{3}

Decimal

286.66666 \dots

Solution steps

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

Expand

(2 x + 4)^{2} : 4 x^{2} + 16 x + 16

= \int_{- 1}^{4} 4 x^{2} + 16 x + 16 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}^{4} 4 x^{2} d x + \int_{- 1}^{4} 16 x d x + \int_{- 1}^{4} 16 d x

\int_{- 1}^{4} 4 x^{2} d x = \frac{260}{3}

\int_{- 1}^{4} 16 x d x = 120

\int_{- 1}^{4} 16 d x = 80

= \frac{260}{3} + 120 + 80

Simplify

\frac{260}{3} + 120 + 80 : \frac{860}{3}

= \frac{860}{3}

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What is the integral from-1 to 4 of (2x+4)^2 ?
The integral from-1 to 4 of (2x+4)^2 is 860/3