What is the integral from 4 to 14 of 16+(4-x)^2 ?

The integral from 4 to 14 of 16+(4-x)^2 is 1480/3

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integral from 4 to 14 of 16+(4-x)^2

\int_{4}^{14} 16 + (4 - x)^{2} d x

\frac{1480}{3}

Decimal Notation

493.33333 \dots

Solution steps

\int_{4}^{14} 16 + (4 - x)^{2} d x

Simplify

16 + (4 - x)^{2} : x^{2} - 8 x + 32

= \int_{4}^{14} x^{2} - 8 x + 32 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_{4}^{14} x^{2} d x - \int_{4}^{14} 8 x d x + \int_{4}^{14} 32 d x

\int_{4}^{14} x^{2} d x = \frac{2680}{3}

\int_{4}^{14} 8 x d x = 720

\int_{4}^{14} 32 d x = 320

= \frac{2680}{3} - 720 + 320

Simplify

\frac{2680}{3} - 720 + 320 : \frac{1480}{3}

= \frac{1480}{3}

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\frac{d}{d x} (\frac{3}{2 x ^{3}})

\frac{d}{d t} (\frac{t}{1 + t ^{2}})

What is the integral from 4 to 14 of 16+(4-x)^2 ?
The integral from 4 to 14 of 16+(4-x)^2 is 1480/3