What is the integral from 1 to 4 of (3-|x-3|) ?

The integral from 1 to 4 of (3-|x-3|) is 13/2

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integral from 1 to 4 of (3-|x-3|)

\int_{1}^{4} (3 - ∣ x - 3 ∣) d x

\frac{13}{2}

Solution steps

\int_{1}^{4} 3 - ∣ x - 3 ∣ d x

Eliminate Absolutes

= \int_{1}^{3} 3 - (- x + 3) d x + \int_{3}^{4} 3 - (x - 3) d x

\int_{1}^{3} 3 - (- x + 3) d x = 4

\int_{3}^{4} 3 - (x - 3) d x = \frac{5}{2}

= 4 + \frac{5}{2}

Simplify

4 + \frac{5}{2} : \frac{13}{2}

= \frac{13}{2}

f (x) = \frac{2 x}{x ^{2} - 7}

\frac{d}{d x} (x^{2} + (y - 1)^{2})

x \to \infty lim (25 - x^{3})

\frac{d}{d x} (\frac{1}{5} sin (x))

y^{^{'''}} + 5 y^{^{''}} - 2 y^{^{'}} - 4 y = 0

What is the integral from 1 to 4 of (3-|x-3|) ?
The integral from 1 to 4 of (3-|x-3|) is 13/2