What is the integral from 1 to 8 of \sqrt[3]{x}-1 ?

The integral from 1 to 8 of \sqrt[3]{x}-1 is 17/4

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integral from 1 to 8 of \sqrt[3]{x}-1

\int_{1}^{8} 3 x - 1 d x

\frac{17}{4}

Decimal

4.25

Solution steps

\int_{1}^{8} 3 x - 1 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_{1}^{8} 3 x d x - \int_{1}^{8} 1 d x

\int_{1}^{8} 3 x d x = \frac{45}{4}

\int_{1}^{8} 1 d x = 7

= \frac{45}{4} - 7

Simplify

\frac{45}{4} - 7 : \frac{17}{4}

= \frac{17}{4}

\int_{1}^{8} 1 + x^{3} d x

\int_{\frac{x}{2}}^{2 x - \frac{x ^{2}}{4}} 1

\int_{1}^{t} 2 x d x

\int_{0}^{s i n (x)} y d y

\int_{0}^{s i n (x)} x d x

What is the integral from 1 to 8 of \sqrt[3]{x}-1 ?
The integral from 1 to 8 of \sqrt[3]{x}-1 is 17/4