What is the integral from 1 to 9 of sqrt(t)(1+t) ?

The integral from 1 to 9 of sqrt(t)(1+t) is 1712/15

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integral from 1 to 9 of sqrt(t)(1+t)

\int_{1}^{9} t (1 + t) d t

\frac{1712}{15}

Decimal

114.13333 \dots

Solution steps

\int_{1}^{9} t (1 + t) d t

Expand

t (1 + t) : t + t^{\frac{3}{2}}

= \int_{1}^{9} t + t^{\frac{3}{2}} d t

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}^{9} t d t + \int_{1}^{9} t^{\frac{3}{2}} d t

\int_{1}^{9} t d t = \frac{52}{3}

\int_{1}^{9} t^{\frac{3}{2}} d t = \frac{484}{5}

= \frac{52}{3} + \frac{484}{5}

Simplify

\frac{52}{3} + \frac{484}{5} : \frac{1712}{15}

= \frac{1712}{15}

What is the integral from 1 to 9 of sqrt(t)(1+t) ?
The integral from 1 to 9 of sqrt(t)(1+t) is 1712/15