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

The integral from 1 to 4 of (1-x)(x-4) is 9/2

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integral from 1 to 4 of (1-x)(x-4)

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

\frac{9}{2}

Decimal Notation

4.5

Solution steps

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

Expand

(1 - x) (x - 4) : 5 x - x^{2} - 4

= \int_{1}^{4} 5 x - x^{2} - 4 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_{1}^{4} 5 x d x - \int_{1}^{4} x^{2} d x - \int_{1}^{4} 4 d x

\int_{1}^{4} 5 x d x = \frac{75}{2}

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 4 d x = 12

= \frac{75}{2} - 21 - 12

Simplify

\frac{75}{2} - 21 - 12 : \frac{9}{2}

= \frac{9}{2}

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

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{( 1 + t ^{2} )}

\int \frac{1}{x} e^{- x} d x

n = 0 \sum \infty cos (5 nπ)

f (x) = ln^{5} (x)

What is the integral from 1 to 4 of (1-x)(x-4) ?
The integral from 1 to 4 of (1-x)(x-4) is 9/2