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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{4 2}{3}

Decimal Notation

1.88561 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{2} u d u

= \int_{0}^{2} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{2}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{2}

Compute the boundaries:

\frac{4 2}{3}

= \frac{4 2}{3}

\frac{d x}{d t} = e^{t}

\int (2 x y - sin (x)) d x

y = x^{3} - 4 x, (- 2, 0)

\int_{1}^{e^{2}} \frac{ln ^{8} ( x ^{2} )}{x} d x

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