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

The integral from 0 to 2 of xsqrt(1+x) is 2/3+(18sqrt(3)-2)/5-2sqrt(3)

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

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

\frac{2}{3} + \frac{18 3 - 2}{5} - 23

Decimal Notation

3.03794 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} (u - 1) u d u

= \int_{1}^{3} (u - 1) u d u

Expand

(u - 1) u : u^{\frac{3}{2}} - u

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{1}^{3} u^{\frac{3}{2}} d u = \frac{18 3 - 2}{5}

\int_{1}^{3} u d u = 23 - \frac{2}{3}

= \frac{18 3 - 2}{5} - (23 - \frac{2}{3})

Simplify

= \frac{2}{3} + \frac{18 3 - 2}{5} - 23

\int_{0}^{\frac{π}{2}} \frac{1}{2}

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

\int_{0}^{1} 2 e^{- t} d t

\int_{- 1}^{7} \frac{x ^{2}}{x ^{3} + 1} d x

\int_{- 2}^{0} 4 - 2^{- x} d x

What is the integral from 0 to 2 of xsqrt(1+x) ?
The integral from 0 to 2 of xsqrt(1+x) is 2/3+(18sqrt(3)-2)/5-2sqrt(3)