What is the integral from 0 to 5 of (6x-x^2)^2-x^2 ?

The integral from 0 to 5 of (6x-x^2)^2-x^2 is 625/3

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integral from 0 to 5 of (6x-x^2)^2-x^2

\int_{0}^{5} (6 x - x^{2})^{2} - x^{2} d x

\frac{625}{3}

Decimal

208.33333 \dots

Solution steps

\int_{0}^{5} (6 x - x^{2})^{2} - x^{2} d x

Expand

(6 x - x^{2})^{2} - x^{2} : x^{4} - 12 x^{3} + 35 x^{2}

= \int_{0}^{5} x^{4} - 12 x^{3} + 35 x^{2} 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_{0}^{5} x^{4} d x - \int_{0}^{5} 12 x^{3} d x + \int_{0}^{5} 35 x^{2} d x

\int_{0}^{5} x^{4} d x = 625

\int_{0}^{5} 12 x^{3} d x = 1875

\int_{0}^{5} 35 x^{2} d x = \frac{4375}{3}

= 625 - 1875 + \frac{4375}{3}

Simplify

625 - 1875 + \frac{4375}{3} : \frac{625}{3}

= \frac{625}{3}

n = 1 \sum \infty \frac{5}{2 + 3 ^{n}}

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\frac{d}{d x} (25 x^{2})

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\frac{d}{d x} (sin (1.5 x))

What is the integral from 0 to 5 of (6x-x^2)^2-x^2 ?
The integral from 0 to 5 of (6x-x^2)^2-x^2 is 625/3