What is the integral from 0 to 5 of (7x-x^2)-(2x) ?

The integral from 0 to 5 of (7x-x^2)-(2x) is 125/6

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integral from 0 to 5 of (7x-x^2)-(2x)

\int_{0}^{5} (7 x - x^{2}) - (2 x) d x

\frac{125}{6}

Decimal Notation

20.83333 \dots

Solution steps

\int_{0}^{5} 7 x - x^{2} - 2 x d x

Simplify

= \int_{0}^{5} - x^{2} + 5 x 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^{2} d x + \int_{0}^{5} 5 x d x

\int_{0}^{5} x^{2} d x = \frac{125}{3}

\int_{0}^{5} 5 x d x = \frac{125}{2}

= - \frac{125}{3} + \frac{125}{2}

Simplify

- \frac{125}{3} + \frac{125}{2} : \frac{125}{6}

= \frac{125}{6}

x \to \infty lim (5 + \frac{6}{x})

n = 0 \sum \infty 4^{- n}

\frac{d}{d y} (y cos (x) + 2 x e^{y})

2 y^{^{''}} - 4 y^{^{'}} + 8 y = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = t

What is the integral from 0 to 5 of (7x-x^2)-(2x) ?
The integral from 0 to 5 of (7x-x^2)-(2x) is 125/6