What is the integral from 0 to 1 of 1+6x^{3/2} ?

The integral from 0 to 1 of 1+6x^{3/2} is 17/5

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integral from 0 to 1 of 1+6x^{3/2}

\int_{0}^{1} 1 + 6 x^{\frac{3}{2}} d x

\frac{17}{5}

Solution steps

\int_{0}^{1} 1 + 6 x^{\frac{3}{2}} d x

Apply the Sum Rule:

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

= \int_{0}^{1} 1 d x + \int_{0}^{1} 6 x^{\frac{3}{2}} d x

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 6 x^{\frac{3}{2}} d x = \frac{12}{5}

= 1 + \frac{12}{5}

Simplify

1 + \frac{12}{5} : \frac{17}{5}

= \frac{17}{5}

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What is the integral from 0 to 1 of 1+6x^{3/2} ?
The integral from 0 to 1 of 1+6x^{3/2} is 17/5