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

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

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

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

- \frac{121}{6}

Decimal Notation

- 20.16666 \dots

Solution steps

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

Expand

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

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

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} u^{2} d u - \int_{0}^{1} u d u - \int_{0}^{1} 20 d u

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

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

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

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

Simplify

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

= - \frac{121}{6}

x = 1, x + y = 6

\frac{e ^{x} - ( 1 + x )}{7 x ^{2}}

\frac{\partial}{\partial x} (e^{- x} \cdot cos (y))

\int x^{- 1} ln (x) d x

x \to - 3 lim (\frac{x + 3}{x + 3})

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