What is the integral from 0 to 1 of (x+4)(x-5) ?

The integral from 0 to 1 of (x+4)(x-5) is -121/6

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integral from 0 to 1 of (x+4)(x-5)

\int_{0}^{1} (x + 4) (x - 5) d x

- \frac{121}{6}

Solution steps

\int_{0}^{1} (x + 4) (x - 5) d x

Expand

(x + 4) (x - 5) : x^{2} - x - 20

= \int_{0}^{1} x^{2} - x - 20 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}^{1} x^{2} d x - \int_{0}^{1} x d x - \int_{0}^{1} 20 d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} 20 d x = 20

= \frac{1}{3} - \frac{1}{2} - 20

Simplify

\frac{1}{3} - \frac{1}{2} - 20 : - \frac{121}{6}

= - \frac{121}{6}

\frac{1.6}{3.2 s + 1}

\int \frac{4 x}{1 + 2 x} d x

x = y^{2}, x = y + 6

\frac{\partial}{\partial s} (s cos (t - \frac{π}{2}))

\frac{d}{d x} (\frac{- 3}{x ^{2} + 9})

What is the integral from 0 to 1 of (x+4)(x-5) ?
The integral from 0 to 1 of (x+4)(x-5) is -121/6