What is the integral from 1 to 6 of wsqrt(w+3) ?

The integral from 1 to 6 of wsqrt(w+3) is 232/5

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integral from 1 to 6 of wsqrt(w+3)

\int_{1}^{6} w w + 3 d w

\frac{232}{5}

Decimal

46.4

Solution steps

\int_{1}^{6} w w + 3 d w

Apply u-substitution

: \int_{4}^{9} (u - 3) u d u

= \int_{4}^{9} (u - 3) u d u

Expand

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

= \int_{4}^{9} u^{\frac{3}{2}} - 3 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_{4}^{9} u^{\frac{3}{2}} d u - \int_{4}^{9} 3 u d u

\int_{4}^{9} u^{\frac{3}{2}} d u = \frac{422}{5}

\int_{4}^{9} 3 u d u = 38

= \frac{422}{5} - 38

Simplify

\frac{422}{5} - 38 : \frac{232}{5}

= \frac{232}{5}

y^{^{''}} + y^{^{'}} + 4 y = 5 x - 7

\int \frac{7}{5} sin (3 x) cos (\frac{x}{3}) d x

\frac{\partial}{\partial x} (\frac{2 y}{x})

3 x x

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

What is the integral from 1 to 6 of wsqrt(w+3) ?
The integral from 1 to 6 of wsqrt(w+3) is 232/5