What is the integral from 0 to 7 of sqrt(1+x) ?

The integral from 0 to 7 of sqrt(1+x) is (32sqrt(2)-2)/3

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integral from 0 to 7 of sqrt(1+x)

\int_{0}^{7} 1 + x d x

\frac{32 2 - 2}{3}

Decimal Notation

14.41827 \dots

Solution steps

\int_{0}^{7} 1 + x d x

Apply u-substitution

: \int_{1}^{8} u d u

= \int_{1}^{8} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{8}

= [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{8}

Compute the boundaries:

\frac{32 2}{3} - \frac{2}{3}

= \frac{32 2}{3} - \frac{2}{3}

Simplify

= \frac{32 2 - 2}{3}

f (x) = 4 x sin (x)

\int cos^{2} (2 x) 2 d x

x \to 0 - lim (\frac{5}{x} - \frac{5}{x})

sin (2 x + 1)

\frac{d}{d x} (\frac{3 x}{2 x + 3})

What is the integral from 0 to 7 of sqrt(1+x) ?
The integral from 0 to 7 of sqrt(1+x) is (32sqrt(2)-2)/3