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

The integral from-1 to 1 of sqrt(x+1)-1 is (4sqrt(2))/3-2

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integral from-1 to 1 of sqrt(x+1)-1

\int_{- 1}^{1} x + 1 - 1 d x

\frac{4 2}{3} - 2

Decimal Notation

- 0.11438 \dots

Solution steps

\int_{- 1}^{1} x + 1 - 1 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}^{1} x + 1 d x - \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} x + 1 d x = \frac{4 2}{3}

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

= \frac{4 2}{3} - 2

\int_{- 3}^{2} ∣ x + 1 ∣ d x

x \to 1 lim (\frac{2 x}{ln ( x )})

(2.9) (9.7)

f (x) = 6 - x

\int_{0}^{2} e^{\frac{x}{2}} d x

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