What is the integral from 0 to 1 of (u+2)(u-3) ?

The integral from 0 to 1 of (u+2)(u-3) is -37/6

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integral from 0 to 1 of (u+2)(u-3)

\int_{0}^{1} (u + 2) (u - 3) d u

- \frac{37}{6}

Decimal Notation

- 6.16666 \dots

Solution steps

\int_{0}^{1} (u + 2) (u - 3) d u

Expand

(u + 2) (u - 3) : u^{2} - u - 6

= \int_{0}^{1} u^{2} - u - 6 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} 6 d u

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

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

\int_{0}^{1} 6 d u = 6

= \frac{1}{3} - \frac{1}{2} - 6

Simplify

\frac{1}{3} - \frac{1}{2} - 6 : - \frac{37}{6}

= - \frac{37}{6}

\int \frac{x ^{2} ln ^{3} ( 1 + x ^{3} )}{1 + x ^{3}} d x

x \to 0 lim (\frac{1}{2 ^{x}})

sin (tan (2 x))

\int 3 x (x^{2} - \frac{1}{3}) d x

f (x) = \frac{x}{x + 1}, at x = - 2

What is the integral from 0 to 1 of (u+2)(u-3) ?
The integral from 0 to 1 of (u+2)(u-3) is -37/6