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

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

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

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

\frac{2 ( 1 + x ) ^{\frac{3}{2}} - 4 2}{3}

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{2}{3} (1 + x)^{\frac{3}{2}} - \frac{4 2}{3}

= \frac{2}{3} (1 + x)^{\frac{3}{2}} - \frac{4 2}{3}

Simplify

= \frac{2 ( 1 + x ) ^{\frac{3}{2}} - 4 2}{3}

\int_{0}^{1} (x^{\frac{1}{2}} - x^{2}) d x

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\int_{\frac{1}{2}}^{1} x^{- 3} - 21 d x

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