What is the integral from 0 to 3 of (xsqrt(1+x)) ?

The integral from 0 to 3 of (xsqrt(1+x)) is 116/15

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integral from 0 to 3 of (xsqrt(1+x))

\int_{0}^{3} (x 1 + x) d x

\frac{116}{15}

Decimal Notation

7.73333 \dots

Solution steps

\int_{0}^{3} x 1 + x d x

Apply u-substitution

: \int_{1}^{4} (u - 1) u d u

= \int_{1}^{4} (u - 1) u d u

Expand

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

= \int_{1}^{4} 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}^{4} u^{\frac{3}{2}} d u - \int_{1}^{4} u d u

\int_{1}^{4} u^{\frac{3}{2}} d u = \frac{62}{5}

\int_{1}^{4} u d u = \frac{14}{3}

= \frac{62}{5} - \frac{14}{3}

Simplify

\frac{62}{5} - \frac{14}{3} : \frac{116}{15}

= \frac{116}{15}

\int_{0}^{r} b \cdot r^{3} + c \cdot r^{2} d r

\int_{0}^{2 π} x^{3} d x

\int_{0}^{2} \frac{π}{2} r d r

\int_{0}^{1} x x - x^{3} d x

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

What is the integral from 0 to 3 of (xsqrt(1+x)) ?
The integral from 0 to 3 of (xsqrt(1+x)) is 116/15