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

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

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

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

\frac{4 2 - 2}{3}

Decimal Notation

1.21895 \dots

Solution steps

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

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}

\frac{d y}{d x} = (x - 3) (y - 10)

\frac{\partial}{\partial x} (x^{2} + 3 x y - y^{2})

x \frac{d y}{d x} + 3 y = \frac{e ^{x}}{x ^{2}}

\int x (x + 4)^{\frac{1}{3}} d x

(2 x - 8)^{4} (x^{2} + x + 1)^{5}

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