What is the integral from 1 to 27 of (4x^{2/3}+1)/(4x^{1/3)} ?

The integral from 1 to 27 of (4x^{2/3}+1)/(4x^{1/3)} is 63

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integral from 1 to 27 of (4x^{2/3}+1)/(4x^{1/3)}

\int_{1}^{27} \frac{4 x ^{\frac{2}{3}} + 1}{4 x ^{\frac{1}{3}}} d x

63

Solution steps

\int_{1}^{27} \frac{4 x ^{\frac{2}{3}} + 1}{4 x ^{\frac{1}{3}}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{1}^{27} \frac{4 x ^{\frac{2}{3}} + 1}{x ^{\frac{1}{3}}} d x

Expand

\frac{4 x ^{\frac{2}{3}} + 1}{x ^{\frac{1}{3}}} : 4 x^{\frac{1}{3}} + \frac{1}{x ^{\frac{1}{3}}}

= \frac{1}{4} \cdot \int_{1}^{27} 4 x^{\frac{1}{3}} + \frac{1}{x ^{\frac{1}{3}}} 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

= \frac{1}{4} (\int_{1}^{27} 4 x^{\frac{1}{3}} d x + \int_{1}^{27} \frac{1}{x ^{\frac{1}{3}}} d x)

\int_{1}^{27} 4 x^{\frac{1}{3}} d x = 240

\int_{1}^{27} \frac{1}{x ^{\frac{1}{3}}} d x = 12

= \frac{1}{4} (240 + 12)

Simplify

\frac{1}{4} (240 + 12) : 63

= 63

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What is the integral from 1 to 27 of (4x^{2/3}+1)/(4x^{1/3)} ?
The integral from 1 to 27 of (4x^{2/3}+1)/(4x^{1/3)} is 63