What is the integral from 1 to 2 of (|x+1|+2) ?

The integral from 1 to 2 of (|x+1|+2) is 9/2

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integral from 1 to 2 of (|x+1|+2)

\int_{1}^{2} (∣ x + 1 ∣ + 2) d x

\frac{9}{2}

Decimal

4.5

Solution steps

\int_{1}^{2} ∣ x + 1 ∣ + 2 d x

Eliminate Absolutes

= \int_{1}^{2} x + 1 + 2 d x

Simplify

= \int_{1}^{2} x + 3 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}^{2} x d x + \int_{1}^{2} 3 d x

\int_{1}^{2} x d x = \frac{3}{2}

\int_{1}^{2} 3 d x = 3

= \frac{3}{2} + 3

Simplify

\frac{3}{2} + 3 : \frac{9}{2}

= \frac{9}{2}

y = \frac{1}{sin ( x )}

\frac{d ^{3}}{d x ^{3}} (\frac{3}{x ^{2}})

30

\frac{d}{d x} (\frac{7}{4 x})

\frac{d}{d x} (\frac{8}{x + 2})

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