What is the integral from 1 to 2 of 1+(x^3)/8 ?

The integral from 1 to 2 of 1+(x^3)/8 is 47/32

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integral from 1 to 2 of 1+(x^3)/8

\int_{1}^{2} 1 + \frac{x ^{3}}{8} d x

\frac{47}{32}

Decimal

1.46875

Solution steps

\int_{1}^{2} 1 + \frac{x ^{3}}{8} 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}^{2} 1 d x + \int_{1}^{2} \frac{x ^{3}}{8} d x

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

\int_{1}^{2} \frac{x ^{3}}{8} d x = \frac{15}{32}

= 1 + \frac{15}{32}

Simplify

1 + \frac{15}{32} : \frac{47}{32}

= \frac{47}{32}

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What is the integral from 1 to 2 of 1+(x^3)/8 ?
The integral from 1 to 2 of 1+(x^3)/8 is 47/32