What is the integral from 1 to 2 of xsqrt(x+3) ?

The integral from 1 to 2 of xsqrt(x+3) is 16/5

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integral from 1 to 2 of xsqrt(x+3)

\int_{1}^{2} x x + 3 d x

\frac{16}{5}

Decimal Notation

3.2

Solution steps

\int_{1}^{2} x x + 3 d x

Apply u-substitution

: \int_{4}^{5} (u - 3) u d u

= \int_{4}^{5} (u - 3) u d u

Expand

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

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

\int_{4}^{5} u^{\frac{3}{2}} d u = 105 - \frac{64}{5}

\int_{4}^{5} 3 u d u = 105 - 16

= 105 - \frac{64}{5} - (105 - 16)

Simplify

105 - \frac{64}{5} - (105 - 16) : \frac{16}{5}

= \frac{16}{5}

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

\int_{4}^{3} (sin (x)) d x

\int_{0}^{1} 2^{x} \cdot 3^{x} d x

\int_{0}^{1} x 33 x^{2} + 16 d x

\int_{0}^{5} 16 - y^{2} d y

What is the integral from 1 to 2 of xsqrt(x+3) ?
The integral from 1 to 2 of xsqrt(x+3) is 16/5