What is the integral from 0 to 2 of (3-x)(x-4) ?

The integral from 0 to 2 of (3-x)(x-4) is -38/3

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integral from 0 to 2 of (3-x)(x-4)

\int_{0}^{2} (3 - x) (x - 4) d x

- \frac{38}{3}

Decimal Notation

- 12.66666 \dots

Solution steps

\int_{0}^{2} (3 - x) (x - 4) d x

Expand

(3 - x) (x - 4) : 7 x - x^{2} - 12

= \int_{0}^{2} 7 x - x^{2} - 12 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}^{2} 7 x d x - \int_{0}^{2} x^{2} d x - \int_{0}^{2} 12 d x

\int_{0}^{2} 7 x d x = 14

\int_{0}^{2} x^{2} d x = \frac{8}{3}

\int_{0}^{2} 12 d x = 24

= 14 - \frac{8}{3} - 24

Simplify

14 - \frac{8}{3} - 24 : - \frac{38}{3}

= - \frac{38}{3}

\int_{0}^{14} x 14 - x d x

\int_{\frac{π}{4}}^{\frac{π}{2}} sec (x) d x

\int_{0}^{1} 100 (4 - 4 y) d y

\int_{0}^{1} 2 t (t + 2) d t

\int_{1}^{\infty} 200 e^{- x} d x

What is the integral from 0 to 2 of (3-x)(x-4) ?
The integral from 0 to 2 of (3-x)(x-4) is -38/3