What is the integral from 0 to 1 of sqrt(2x+1) ?

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

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integral from 0 to 1 of sqrt(2x+1)

\int_{0}^{1} 2 x + 1 d x

\frac{3 3 - 1}{3}

Decimal Notation

1.39871 \dots

Solution steps

\int_{0}^{1} 2 x + 1 d x

Apply u-substitution

: \int_{1}^{3} \frac{u}{2} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{1}^{3} u d u

Apply the Power Rule

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

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

Compute the boundaries:

23 - \frac{2}{3}

= \frac{1}{2} (23 - \frac{2}{3})

Simplify

= \frac{3 3 - 1}{3}

x \to \infty + lim (25 x^{4} + 9 x - 5 x^{2})

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

\int \frac{e ^{x}}{2} + x x d x

f (x) = - 3 cos (x), at x = - \frac{π}{3}

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

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