What is the integral from 0 to 1 of sqrt(1+x) ?

The integral from 0 to 1 of sqrt(1+x) is (4sqrt(2)-2)/3

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integral from 0 to 1 of sqrt(1+x)

\int_{0}^{1} 1 + x d x

\frac{4 2 - 2}{3}

Decimal Notation

1.21895 \dots

Solution steps

\int_{0}^{1} 1 + x d x

Apply u-substitution

: \int_{1}^{2} u d u

= \int_{1}^{2} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{2}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{2}

Compute the boundaries:

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

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

Simplify

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

\int_{0}^{1} π (1 - x^{2})^{2} d x

(3 + 4 x) + (5 y - 1) \frac{d y}{d x} = 0

f (x) = x^{2} + 5 x - 2, at x = 4

x \to 0 lim (2 x^{2} + 3 x)

\int \frac{1 + 4 x}{1 + x + 2 x ^{2}} d x

What is the integral from 0 to 1 of sqrt(1+x) ?
The integral from 0 to 1 of sqrt(1+x) is (4sqrt(2)-2)/3